MTH101 Mega File Final Term PDF

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CoNtaCt us oNly What’s app 0302-1400084  Calculus And Analytical Geometry MTH 101 Virtual University of Pakistan Knowledge beyond the boundaries 45-Planning Production Levels: Linear Programming VU TABLE OF CONTENTS : Lesson 1 :Coordinates, Graphs, Lines 3 Lesson 2 :Absolute Value 15 Lesson 3 :Coordinate Planes and Graphs 24 Lesson 4 :Lines 34 Lesson 5 :Distance; Circles, Quadratic Equations 45 Lesson 6 :Functions and Limits 57 Lesson 7 :Operations on Functions 63 Lesson 8 :Graphing Functions 69 Lesson 9 :Limits (Intuitive Introduction) 76 Lesson 10:Limits (Computational Techniques) 84 Lesson 11: Limits (Rigorous Approach) 93 Lesson 12 :Continuity 97 Lesson 13 :Limits and Continuity of Trigonometric Functions 104 Lesson 14 :Tangent Lines, Rates of Change 110 Lesson 15 :The Derivative 115 Lesson 16 :Techniques of Differentiation 123 Lesson 17 :Derivatives of Trigonometric Function 128 Lesson 18 :The chain Rule 132 Lesson 19 :Implicit Differentiation 136 Lesson 20 :Derivative of Logarithmic and Exponential Functions 139 Lesson 21 :Applications of Differentiation 145 Lesson 22 :Relative Extrema 151 Lesson 23 :Maximum and Minimum Values of Functions 158 Lesson 24 :Newton’s Method, Rolle’s Theorem and Mean Value Theorem 164 Lesson 25 :Integrations 169 Lesson 26 :Integration by Substitution 174 Lesson 27 :Sigma Notation 179 Lesson 28 :Area as Limit 183 Lesson 29 :Definite Integral 191 Lesson 30 :First Fundamental Theorem of Calculus 200 Lesson 31 :Evaluating Definite Integral by Subsitution 206 Lesson 32 :Second Fundamental Theorem of Calculus 210 Lesson 33 :Application of Definite Integral 214 Lesson 34 :Volume by slicing; Disks and Washers 221 Lesson 35 :Volume by Cylindrical Shells 230 Lesson 36 :Length of Plane Curves 237 Lesson 37 :Area of Surface of Revolution 240 Lesson 38:Work and Definite Integral 245 Lesson 39 :Improper Integral 252 Lesson 40 :L’Hopital’s Rule 258 Lesson 41 :Sequence 265 Lesson 42 :Infinite Series 276 Lesson 43 :Additional Convergence tests 285 Lesson 44 :Alternating Series; Conditional Convergence 290 Lesson 45 :Taylor and Maclaurin Series 296 23-Maximum&Minimum Values of Functions VU Lecture # 23 Maximum and Minimum Values of Functions Absolute Extrema Finding Absolute Extrema for a continuous function Summary of extreme behaviors of functions over Applied maximum and minimum problems Problems involving finite closed intervals Problems involving intervals that are non-finite and closed Absolute Extrema Previously we talked about relative maxima, relative minima of functions. These were like the highest mountain and the deepest valley in a given vicinity or neighborhood. Now we will talk about absolute maximum and minimum values of functions. These are like the highest peak in a mountain range, and the deepest valley. Absolute maximum means that the value is the maximum one over the entire domain of the function. Absolute minimum means that the value is the minimum one over the entire domain of the function. If we think of the Earth’s surface as defining some function, then its absolute maximum will be Mt. Everest, and absolute minimum will be the Marianna Trench in the Pacific Ocean near Hawaii. Example Consider the following picture of the graph of f (x) = 2x+1 on the interval [0,3). The minimum value is 1 at x = 0. But is there a maximum? © Copyright Virtual University of Pakistan 158 23-Maximum&Minimum Values of Functions VU No. Because the function is defined on the interval [0,3) which excludes the point x = 3. So note that you can get very close to 7 as the maximum value as you get very close to x = 3, but this is in a limiting process, and you can always get more closer to 7, and yet never EQUAL 7! So 7 looks like a max, but its NOT! The question of interest given a function f(x) is: Does f (x) has a maximum (minimum) value? If f (x) has a maximum (minimum) , what is it? If f (x) has a maximum value, where does it occur? Won’t prove this as its difficult. Just use it. This theorem doesn’t tell us what the max and the min are, just the conditions on a function which will make it have a max or min Example Note that in the previous example we saw a function f (x) = 2x+1 which was defined over the interval [0, 3). This one is a continuous function on that interval, but had no maximum because the interval was not closed! Example Here is another function graphed over the interval [1,9]. Although the interval is closed, the function not continuous on this interval as we can see from the graph, and so has no maximum or minimum values on that interval. © Copyright Virtual University of Pakistan 159 23-Maximum&Minimum Values of Functions VU Step 1: Find the critical points of f in (a,b) Step 2: Evaluate f at all the critical points and the endpoint a and b Step 3: The largest of the values in Step 2 is the maximum value of f on [a,b] and the smallest is the minimum. Example Find the maximum and the minimum values of f ( x) =2 x 3 − 15 x 2 + 36 x on the interval [1, 5] Since f is a polynomial, it’s continuous and differentiable on the interval (1, 5) © Copyright Virtual University of Pakistan 160 23-Maximum&Minimum Values of Functions VU f / ( x) = 6 x 2 − 30 x + 36 = 0 f / ( x) = ( x − 3)( x − 2) = 0 So f / is 0 at x = 2 and x = 3. Max or min will occur at these two points or at the end points. Evaluate the function at the critical points and the endpoints and we see that max is 55 at x = 5 and min is 23 at x = 1 We want to know the max and min over (.−∞, +∞) Here is how to find them for continuous functions Summary of extreme behaviors of functions over (a,b) Find the max and min values if any, of the function f ( x) =x 4 + 2 x 3 − 1 on the interval (−∞, +∞) This is a continuous function on the given interval and lim ( x 4 + 2 x3 − 1) = +∞ and lim ( x 4 + 2 x3 − 1) = +∞ x →+∞ x →−∞ © Copyright Virtual University of Pakistan 161 23-Maximum&Minimum Values of Functions VU So f has a minimum but no maximum on (-inf, +inf). By Theorem 4.6.5, the min must occur at a critical point. So f '( x) = 4 x3 + 6 x 2 = 2 x 2 (2 x + 3) = 0 This gives x = 0, and x = -3/2 as the critical points. Evaluating gives min = -43/16 at x = -3/2 Applied maximum and minimum problems We will use what we have learnt so far to do some applied problems in OPTIMIZATION. Optimization is the way efficiency is got in business, machines and even in nature in terms of animals competing for resources. Problems involving continuous functions and Finite closed intervals These are problems where the function is defined over a closed interval. These problems always have a solution because it is guaranteed by the extreme values theorem. Example Find the dimensions of a rectangle with perimeter 100 ft whose area is as large as possible. Let x = length of the rectangle in feet y = width of the rectangle in feet A = area of the rectangle Then A = xy We want to maximize the area A = x(50 − x) = 50 x − x 2 Perimeter = 100ft = 2x + 2y or y = 5 – x Use this values of y in the equation A = xy to get A function of x Because x represents a length, it cannot be negative and it cannot be a value that exceeds the perimeter of 100 ft. So we have the following constraints on x 0 ≤ x ≤ 50 So the question is of finding the max of = A 50 x − x 2 on the interval [0,50]. By what we have seen so far, that max must occur at the end points of this interval or at a critical point Critical points: dA = 50 − 2 x = 0 ⇒ x = 25 dx So now we substitute x = 0, x = 25, and x = 50 into the function A to get the max 625 at the point x = 25. Note that y = 25 also for x = 25. So the rectangle with perimeter 100 with the greatest area is a square with sides 25ft. © Copyright Virtual University of Pakistan 162 23-Maximum&Minimum Values of Functions VU Example An open box is to be made from a 16 inch by 30 inch piece of cardboard by cutting out squares of equal size from the 4 corners and bending up the sides. What size should the squares be to obtain a box with largest possible volume? If we cut out the squares from the 4 corners of the cardboard, the resulting BOX will have dimensions (16 – 2x) by (30 – 2x) s So want to find max of the function on [0,8]. dV 10 = 480 − 184 x + 12 x 2 = 0 ⇒ x= and x = 12 dx 3 Because x = 12 is out of [0, 8], ignore it. Check V at the end points and at x = 10/3. We see then that V = 19600/27 is max when x = 10/3. © Copyright Virtual University of Pakistan 163 24-Newton’s Method, Rolle’s & Mean Value Theorem VU Lecture # 24 Newton’s Method, Rolle’s Theorem, and the Mean Value Theorem Newton’s method for approximating solutions to f(x)=0 Some difficulties with Newton’s method Rolle’s theorem Mean Value Theorem Newton’s method for approximating solutions to f (x) =0 b We have seen in algebra that the solution for the equation ax + b =0 is x = −. a Similarly we have algebraic formulas for polynomial equation up to degree 5. But there is no algebraic solution to the equation of the kind x − cos( x) = 0 For this equation and many like it, we settle for approximate solutions. How do we approximate the solutions? There are many methods, and one of them is Newton’s method. Here is how Newton's method works to find approximate solutions to equations. What does it mean for an equation to have a solution? or that f (x) = 0 ? It means that we are looking for those x values, for which the corresponding y value or f (x) is 0. This means that the solutions are those points where the graph of the function crosses the x axis. Suppose that x = r is the solution we are looking for. Let's approximate it by an initial guess called. We draw a line tangent to the graph of the given function at the point. If the tangent line is not parallel to the x-axis, then it will x2 eventually intersect the x-axis at some point x1 which will generally be closer to r than We will repeat the process, with a tangent line at x2 that meets x axis at x3, and so on… This is Newton’s method. We need a formula for this method. Note that the tangent line at x1 has the equation y − f ( x= 1) f '( x1 )( x2 − x1 ) If f '( x1 ) ≠ 0 , then the line meets the x axis at ( x2 , 0) Plug this coordinate into above equation, we get − f ( x= 1) f '( x1 )( x2 − x1 ) f ( x1 ) ⇒ x2 = x1 − f '( x1 ) Repeating this process for a third point ( x3 , 0) gives − f ( x= 2) f '( x2 )( x3 − x2 ) f ( x2 ) ⇒ x3 = x2 − f '( x2 ) © Copyright Virtual University of Pakistan 164 24-Newton’s Method, Rolle’s & Mean Value Theorem VU In general, then we have f ( xn ) xn += 1 xn − f '( xn ) There is limiting process involved here for finding the solutions. We get as close as we like to the solution. Example The equation x = cos( x) has a solution between 0 and 1. Approximate it using Newton’s method. Rewrite as x = cos( x) so our function is f ( x)= x − cos( x). The derivative is f / ( x) = 1 + sin( x) So we have xn − cos( xn ) xn += xn − 1 + sin( xn ) 1 As our approximation formula Here is a graph of the situation. From the graph it looks like the solution is closer to 1 than 0. So we will use x 1 = 1.So we get the following approximations. x − cos( x1 ) 1 − cos(1) x2 = x1 − 1 = 1− = 0.7503 1 + sin( x1 ) 1 + sin(1) x − cos( x2 ) 0.7503 − cos(0.7503) x3 = x2 − 2 = 0.7503 − = 0.7391 1 + sin( x2 ) 1 + sin(0.7503) You may continue if you will, but we will say that the solution is approximately x ≈ 0.7391 Some difficulties with Newton’s method Newton’s method does not always work. If for some values of n f / ( xn ) then the formula for Newton’s method involves division by 0 and we are out of business. © Copyright Virtual University of Pakistan 165 24-Newton’s Method, Rolle’s & Mean Value Theorem VU Such a case will occur if the tangent line for some approximation has slope 0 or is parallel to the x axis. Sometimes the approximations don’t converge to a solution. 1 Consider the equation x 3 = 0 The only solution is x = 0. Let's approximate it by Newton’s Method with initial approx x 1 = 1. We get the following Formula 1 xn +1 = x − ( xn ) 3 = −2 xn n 2 1 ( xn ) 3 − 3 Plug in x 1 =1 and then the following approx to see that the values do not converge Rolle’s theorem Rolle’s Theorem says essentially that for a certain kind of function, if it crosses the x-axis at two point, then there is one point between those two points where the derivative of f is 0. Example The function f ( x) = sin( x) is continuous and differentiable everywhere, hence continuous on [0 , 2π] and differentiable on (0, 2π). Also, f (0) = sin (0) = 0 and f(2π)=sin (2π) = 0 So the function satisfies the hypotheses of Rolle’s Theorem. So there exists a point c in the interval (0 , 2π) such that = = f / (c) cos( c) 0 f (b) − f (a ) y= − f (a) ( x − a) b−a ⇒ f (b) − f (a ) =y ( x − a) + f (a) b−a Here is a more tangible way to think of Roll’s theorem © Copyright Virtual University of Pakistan 166  24-Newton’s Method, Rolle’s & Mean Value Theorem VU I leave from Lahore to Islamabad. When i start driving from Lahore, my velocity is 0, and when i reach Islamabad, my velocity is 0 as well. Velocity is a continuous function on the interval [0 , 376]. Also, velocity is differentiable on (0 , 376) as its derivative acceleration is defined at each point on the velocity curve. Hence during my drive from Lahore to ISB, there is some point on the motorway where the acceleration of the car was 0. One could argue: What if you keep accelerating on the motorway! Mean Value Theorem This says basically that under the right conditions, a function will have the same slope for the tangent line at a point as that of a certain secant line. Proof of MVT From this figure we have the following: Slope of Secant line joining A and B:  f (b) − f (a )  v( x)= f ( x) −  ( x − a) + f (a)   b−a  Since f(x) is continuous [a , b] and differentiable on (a , b), so is v(x) by its formula involving f(x). Also note that =v(a ) 0= and v(b) 0 So v(x) satisfies the assumptions of Roll’s theorem on the interval [a , b]. So there is a point c in (a , b) such that v / ( c)=0. But note that © Copyright Virtual University of Pakistan 167 24-Newton’s Method, Rolle’s & Mean Value Theorem VU  f (b) − f (a )  v= / ( x) f / ( x) −    b−a  f (b) − f (a )  ⇒ v / (c ) = f / (c ) −    b−a So by this last formula, at the point where v / (c) = 0 , we have f (b) − f (a ) f / (c ) = b−a © Copyright Virtual University of Pakistan 168 25-Integration VU Lecture # 25 Integrations In this lecture we will look at the beginnings of the other major Calculus problem. The Area Problem Anti-derivatives (Integration) Integration formulas Indefinite Integral Properties of Indefinite Integral The Area Problem Given a continuous and non negative function on an interval [a , b], find the area between the graph of f and the interval [a , b] on the x-axis. Instead of trying to solve a particular case like the one in the picture we just saw, we will generalize to solve this problem where the right end point will be any number x greater than or equal to b instead of just b. We will denote the area we are trying to find as A(x) because this will be a function of x now as it depends on how far away x is from a. It was the idea of Newton and Leibniz that to find the unknown area A(x), first find it derivative A/ ( x) and use this derivative to determine what A (x) is ! Interesting approach. So we want to find out first. A( x + h) − A( x) A '( x) = lim h →0 h Let’s assume for now that h > 0 The top of the derivative quotient is the difference of the two areas A(x) and A(x + h) Let c be the midpoint of between x and x +h. Then the difference of areas can be approximated by the area of the rectangle with base length h and height f ( c). © Copyright Virtual University of Pakistan 169  25-Integration VU So we have A( x + h) − A( x) f (c) ⋅ h ≈ = f (c ) h h Note that the error in the approximation from this rectangle in A will approach 0 as h goes to 0. Then we have A( x + h) − A( x) = = lim f (c) A/ ( x) lim h →0 h h →0 As h goes to 0, c approaches x. Also, f is assumed to be a continuous function, so we have that f( c) goes to f (x) as c goes to x. Thus lim f (c) =f ( x) ⇒ A/ ( x) =f ( x) h →0 So: The derivative of the area function A (x) is the function whose graph forms the upper boundary of the region under which the area is to be found Example Find the area of the region under the graph of= y f= ( x) x 2 over the interval [0 , 1 ] Look at the situation over the interval [0 , x]. Then we have from the discussion that A '( x) = x 2 To find A(x) we look for a function whose derivative is x 2 This is called an antidifferentiation problem as we are trying to find A(x) by undoing a differentiation. A guess is the function 1 A( x) = x3 3 This is a formula for the areas function. So on the interval [0 , 1], we have x = 1 and our result is A(1) = 1/3 units Anti-derivatives Definition 5.2.1 A function F is called antiderivative of a function f on a given interval if F/(x) = f (x) for all x in the interval. © Copyright Virtual University of Pakistan 170  25-Integration VU Example 1 3 1 3 1 3 The functions x, x −π, x + C are all anti-derivatives of 3 3 3 =f ( x) x 2 on the interval (−∞, +∞) As the derivative of each is f ( x) = x 2 If F (x) is any anti-derivative of f (x), then so is F (x) +C where C is a constant. Here is a theorem Indefinite Integral The process of finding anti-derivatives is called anti-differentiation or Integrations. d If there is some function F such that [ F ( x)] = f ( x) dx Then function of the form F (x) +C are anti-derivatives of f (x). We denote this by ∫ f ( x= )dx F ( x) + C The symbol ∫ is called the integral sign and f (x) is called the integrand. It is read as the “Indefinite integral of f (x) equals F (x)” ∫ f ( x= )dx F ( x) + C The right side of the above equation is not a specific function but a whole set of possible functions. That’s why we call it the Indefinite integral. C is called the constant of integrations. Example As we saw earlier, the anti-derivatives of f ( x) = x 2 are functions of the form So we can write 1 3 F= ( x) x +C 3 The dx serves to identify the independent variable in the function involved in the integration. © Copyright Virtual University of Pakistan 171 25-Integration VU Examples: Example From the table we just saw, we obtain the following results. x3 ∫ x dx= +C 2 3 x4 ∫ = +C 3 x dx 4 1 1 ∫ x5 dx =∫ x dx = −5 − 4 +C 4x Properties of Indefinite Integral d  f ( x)dx  = f ( x) dx  ∫  © Copyright Virtual University of Pakistan 172  25-Integration VU Example Evaluate ∫ 4 cos( x)dx ∫ 4 cos( x)= dx 4 ∫ cos( x)= dx 4[sin( x) += C ] 4sin( x) + K Where 4C = K Example ∫ (x + x)dx 2 1 3 1 2 ∫ (x + x)dx = ∫ x dx + ∫ x ⋅ dx = x + x +C 2 2 3 2 Generalized version of the Theorem 5.2.3 b and c ∫ [c f ( x) + c f ( x) +... + c f ( x)]dx 1 1 2 2 n n = c ∫ f ( x)dx +c ∫ f ( x)dx +... + c ∫ f 1 1 2 2 n n ( x)dx Example ∫ (3x − 2 x + 7 x + 1)dx 6 2 = 3∫ x dx − 2 ∫ x dx + 7 ∫ xdx + ∫ dx 6 2 3x 7 2 x3 7 x 2 = − + + x+C 7 3 2 Example cos( x) ∫ sin 2 ( x) dx 1 cos( x) =∫ dx. sin( x) sin( x) ∫ cos ec( x) cot( x)dx = = − cos ec( x) + c © Copyright Virtual University of Pakistan 173 26-Integration by substitution VU Lecture # 26 Integration by substitution This is like the Chain Rule we saw for differentiation. The idea is to integrate functions that are composition of different functions. d [G (u )] = f (u ) du Also this means that d ∫ f (u= )du ) ] du G (u ) + C ∫ du [G(u= Let U be a function of x. Then we have d d du du [G= (u ) ] [G (u )]= ⋅ f (u ) dx du dx dx Now if we apply the integral on both sides w. r. t x we get  du  d dx ∫ [G (u ) ]= ∫  f (u ) dx = dx G (u ) + C dx ⇒  du  ∫  f (u )  dx = ∫ f (u )du dx  Example ∫ (x + 1)50.2 x dx 2 du Let = u x + 1. Then = 2x. 2 dx Now we can rewrite the given problem as  du  ∫ (x + 1)50.2 x dx = ∫ u = ∫u 2 50 50 dx du dx   du  ∫  f (u ) dx  dx = ∫ f (u )du © Copyright Virtual University of Pakistan 174 26-Integration by substitution VU u 51 ( x 2 + 1)51 ∫ u du = 51 + C = 51 + C , where = u x2 + 1 50 Caution: Don’t feel tempted to just add 1 to the power 50 in the original problem!! That will be incorrect. The reason is that the “Power Rule” for the integrals is applicable to function which are not a composition of others. In this example, ( x 2 + 1)50 is a composition of two functions. Here is a summary of the general procedure we need to follow to do integration by u -substitution. Step 1: Make the choice for u, say u = g (x) Step 2: Compute du/dx = g/(x) Step 3: Make the substitution u = g (x), du = g/(x) dx in the original integral. By this point, the whole original integral should be in terms of u and there should be no x’s in it. Step 4: Evaluate the resulting integral Step 5: Replace u by g (x) so the final answer is in x. No hard and fast rule for choosing u. The choice of u should be such that the resulting calculus Results in a simplified integral. There may be more than one choice of u in some cases. Only practice and experience makes things easier. Think of this choice of u like playing chess. You need to choose it so that the future looks bright for problem solution! Integrand is the derivative of a known function with a constant added or subtracted from the independent variable. If the integrand is the derivative of a known function with a constant added or subtracted from the independent variable, then the substitution is the easiest. Here is table 5.2.1 that we saw earlier. This will help recall BASIC integration formulas. © Copyright Virtual University of Pakistan 175 26-Integration by substitution VU We want to integrate a function in which “The integrand is the derivative of a known function with a constant added or subtracted from the independent variable” © Copyright Virtual University of Pakistan 176 26-Integration by substitution VU Example ∫ sin( x + 9)dx =∫ sin(u )du =−cos(u ) + C =−cos( x + 9) + C u= x + 9 du =1 ⇒ du = 1.dx dx The integrand sine is the derivative of cosine function Example u 24 ( x − 8) 24 ∫ ( x − 8) dx = ∫ u du = +C= +C 23 23 24 24 u= x − 8 = du 1.=dx dx Integrand is the derivative of a known function and a constant multiplies the independent variable Example du 1 1 1 ∫ cos(5 x= )dx ∫ cos(u = ) 5 5∫ cos(u=) du 5 sin(u ) = + C 5 sin(5 x) + C u = 5x du du = 5dx ⇒ = dx 5 Example ∫ sin 2 ( x)cos( x)dx Let u = sin( x), then du = cos( x)dx So © Copyright Virtual University of Pakistan 177 26-Integration by substitution VU u3 sin 3 ( x) ∫ sin ( x)cos( x)dx = ∫ u du = 3 + C = 3 + C 2 2 Example cos x ∫ x dx Let u = x , then 1 du = dx 2 x So cos x ∫ x =dx ∫ 2cos(u= )du 2 ∫ cos(u= )du 2sin(u ) = + C 2sin( x ) + C Complicated example  du  ∫t 3 − 5t 5 dt ∫ ( x 2 + 1)50.2 x dx = ∫ u = ∫u 4 50 50 dx du dx  What should the substitution be? Well the idea is to get an expression to equal u so that when we differentiate it, we get a formula that involves a du and everything that was left over in x after the substitution. ∫t 3 − 5t 5 dt 4 u= 3 − 5t 5 1 du =−25t 4 dt ⇒ − du =t 4 dt 25 So 4 1 3 1 1 1 u ∫t − ∫ 3 u du = 3 − 5t 5 dt = − ∫ u du = − +C 4 3 25 25 25 4 3 4 3 = − (3 − 5t ) + C 5 3 100 © Copyright Virtual University of Pakistan 178 27-Sigma Notation VU Lecture #. 27 Sigma Notation Sigma notation is used to write lengthy sums in compact form. Sigma or Σ is an Upper case letter in Greek. This symbol is called Sigma or summation as it is used to represent lengthy sums. Here is an example of how this notation works. Consider the sum 12 + 22 + 32 + 42 + 52 Note that every term in this sum is the square of an integer from 1 to 5. Let's assign these integers a variable k and keep it in mind that this k can take on values from 1 to 5. Then we can say that k^2 represent each of the elements in the sum. So we can write 5 this as ∑k k =1 2 “Summation of k2 where k goes from 1 to 5 Example 8 ∑k k =4 3 = 43 + 53 + 63 + 73 + 83 5 ∑ 2k = k =1 2(1) + 2(2) + 2(3) + 2(4) + 2(5) = 2 + 4 + 6 + 8 + 10 5 ∑ (−1) (2k + 1) =1 − 3 + 5 − 7 + 9 − 11 k =0 k The number on the top are called the upper limits of the summation and the numbers at the bottom are called lower limits of the summation. The letter k is called the index of the summation It is not necessary that the letter k represents the index of summation. We could use i or j or m etc. Example 4 1 4 1 4 1 ∑i, ∑ =i 1 = ,∑ , n j1 j n 1= All denote the sum © Copyright Virtual University of Pakistan 179 27-Sigma Notation VU 1 1 1 1+ + + 2 3 4 If the upper and lower limits are the same, then the summation reduces to just one 2 term ∑k k =2 3 = 23 If the expression to the right of the summation does not involve the index of the summation, then do the following 5 ∑2 = 2 + 2 + 2 + 2 + 2 i =1 6 ∑x k =3 3 = x3 + x3 + x3 + x3 A sum can be written in more than one way with the Sigma notation if we change the limits of the summation Example The following summations all represent the sum of the first five positive integers 5 ∑ 2k = 2 + 4 + 6 + 8 + 10 k =1 4 ∑ (2k + 2) = 2 + 4 + 6 + 8 + 10 k =0 6 ∑ (2k − 2) = 2 + 4 + 6 + 8 + 10 k =2 Changing the index of the summation It is often necessary and useful to change a given Sigma notation for a sum to another sigma notation with different limits of summation Example 7 Express ∑5 k =3 k −2 in sigma notation so that the lower limit is 0 rather than 3. Define a new summation index j by the following formula j = k-3 => k = j+3 Then as k runs from 3 to 7, j runs from 0 to 4. So © Copyright Virtual University of Pakistan 180 27-Sigma Notation VU 7 4 4 = ∑ 5k − 2 = k 3 =j 0 ∑= j ( j +3)−2 =j 0 ∑j j +1 You should check that the two actually represent the same sum by putting values into the index in both notations. To represent a general sum, we will use letters with subscripts. Example a1 + a2 + a3 represent the general sum with three terms This can also be written as in sigma notation 3 ∑a k =1 k =a1 + a2 + a3 General sum with n terms can be written as n ∑b k =1 k =b1 + b2 + b3 +... + bn Properties of Sigma notation Here are a few properties of sigma notation that will be helpful later on Here are some sum formulas written in sigma notation that will be helpful later. © Copyright Virtual University of Pakistan 181 27-Sigma Notation VU Example 30 30 30 30 =k 1 ∑ k (k + 1)= ∑ (k =k 1 2 + k= = ) k 1= ∑k + ∑k k 1 2 (30)(31)(61) 30(31) = + = 9920 6 2 Theorem 5.4.2 a) and b) In a formula involving summation like this one n n(n + 1)(2n + 2) ∑k k =1 2 = 6 n(n + 1)(2n + 2) ⇒ 12 + 22 +... + n 2 = 6 The left part is called the open form of the sum. The right part is called the closed form of the sum. © Copyright Virtual University of Pakistan 182 28-Area as Limits VU Lecture # 28 Area as Limits Definition of Area Some technical considerations Numerical approx of area Look at the figure below In this figure, there is a region bounded below by the x-axis, on the sides by the lines x= a and x = b, and above by a curve or the graph of a continuous function y = f(x) which is also non-negative on the interval [a, b]. Earlier we saw that such an area can be computed using anti-derivatives. Let's make the concept precise. Recall that the slope of the tangent line was defined in terms of the limit of the slopes of secant lines. Similarly, we will define area of a region R as limits of the areas of simpler regions whose areas are known. We will break up R into rectangles, and then find the area of each rectangle in R and then add all these areas up. The result will be an approximation to the region R. Let's call it If we let n go to infinity, the resulting will give a better and better approximation to R as the rectangle in R will get thinner and thinner, and the gaps will be filled in. Here is the formal idea Choose an arbitrary positive integer n, and divide the interval [a,b] into n subintervals of width b−a by inserting n-1 equally spaced points between a and b say x1 , x2 ,...xn−1 n © Copyright Virtual University of Pakistan 183 28-Area as Limits VU These points of subdivision form a regular partition of [a,b] a, x1 , x2 ,...xn−1 , b Next draw a vertical line through the points x1* , x2* ,...xn* This will divide the region R into n strips of uniform width Now we want to approximate the area of each strip by the area of a rectangle. For this, choose an arbitrary point in each subinterval , Over each subinterval construct a rectangle whose height is the values of the function f at the selected arbitrary point. The union of these rectangles form the region Rn that we can regard as a reasonable © Copyright Virtual University of Pakistan 184 28-Area as Limits VU approx to the region R. The area of Rn can be got by adding the areas of all the rectangles forming it If now we let n get big, the number of rectangles gets big, and the gaps btw the curve and the rectangles get filled in. As n goes to infinity, the approx gets as good as the real thing. = So we define A = area ( R) lim [area ( Rn )] n→+∞ For all the following work and computation, we will treat n as a POSITIVE integer For computational purposes, we can write A in a different form as follows b−a In the interval [a,b], each of the approximation rectangles has width n b−a We call this delta x or ∆x = n The heights of the approximating rectangles are at the points x1* , x2* ,...xn* So the approximating rectangles making up the region have areas f ( x1* )∆x, f ( x2* )∆x,..., f ( xn* )∆x © Copyright Virtual University of Pakistan 185 28-Area as Limits VU So the total area of Rn is given by = area ( Rn) f ( x1* )∆x + f ( x2* )∆x +... + f ( xn* )∆x OR n = area ( Rn ) ∑ f (x k =1 k * )∆x So now A can be written as n =A lim ∑ f ( xk * )∆x n→+∞ k =1 This we will take to be the PRECISE def of the area of the region R Some Technical Considerations The points x1* , x2* ,...xn* were chosen arbitrarily. What if some other points were chosen? Would the resulting values for f(x) at these points be different? And if so then the definition of area will not be well-defined? It is proved in advanced courses that since f is continuous it doesn’t matter what points are taken in a subinterval. Usually the point xk* in a subinterval is chosen so that it is the left end point of the interval, the right end point of the interval or the midpoint of the interval. Note that we divided the interval [a,b] by the points x1 , x2 ,..., xn−1 with x0 = a and xn = b into equal parts of width ∆x © Copyright Virtual University of Pakistan 186 28-Area as Limits VU SO xk * = xk −1 = a + (k − 1)∆x Left end point xk * = xk = a + k ∆x Right end point 1 1 xk * = ( xk −1 + xk ) =a + (k − )∆x Midpoint 2 2 Example Use the definition of the Area with xk * as the right end point of each subinterval to find the area under the line y = x over the interval [1,2]. Subdivide [1,2] into n equal parts, then each part will have length b − a 2 −1 1 ∆= x = = * We are taking xk as the right end point so we n n n k Have xk* = a + k ∆x = 1 + n Thus, the kth rectangle has area © Copyright Virtual University of Pakistan 187  28-Area as Limits VU  k  k 1 f ( xk * )∆x = xk *∆x = 1 +  ∆x = 1 +   n  nn And the sum of the areas of the n rectangles is n 3 1  3 =A lim ∑ f ( xk= * )∆x lim  +=  n→+∞ k =1  n→+∞ 2 2n  2 Note that the region we have computed the area of is a trapezoid with height h = 1 and bases b 1 = 1 and b 2 = 2. From basic geometry we have that 1 1 3 Area of trapezoid = = A h(b1 + b2= ) (1)(1 + 2) = 2 2 2 Example Same problem as before but with left end points.[ 1,2]. Use the definition of the Area with xk* as the left end point of each subinterval to find the area under the line y = x over the interval [1,2]. Subdivide [1,2] into n equal parts, then each part will have length b − a 2 −1 1 ∆= x = = n n n We are taking xk * as the left end point so we k Have xk * = a + k ∆x = 1 + n Thus, the kth rectangle has area  k  k 1 f ( xk * )∆x = xk *∆x = 1 +  ∆x = 1 +   n  nn And the sum of the areas of the n rectangles is n  k  1  n  1 k   1 n n 1 n = ∑ k 1 = ∆x ∑ 1 + = f ( xk )= * k 1  n = ∑ + =  n  k 1  n n 2 = ∑1 + n2 ∑  n k 1= k 1 k 1 1 1  = ⋅ n + 2  n(n + 1)  n n 2  Note that the region we have computed the area of is a trapezoid with height h = 1 and bases b1 = 1 and b2 = 2. From basic geometry we have that © Copyright Virtual University of Pakistan 188 28-Area as Limits VU 1 1 3 Area of trapezoid = A = h(b1 + b= 2) (1)(1 + 2) = 2 2 2 Example Use the area definition with right endpoints of each subinterval to find the area under the parabola y= 9 − x 2 over the interval [0,3]. Do calculations from book on page 272 (end) and page 273 for example. Numerical Approximations of Area As we have all seen so far, the computations involved in computing the limits are tedious and long. In some cases, it is even impossible to carry out the computations by the definition of the area. In such cases, it is easier to get a GOOD approx for the area using large values of n and using a computer or a calculator. Example Use a computer of a calculator to find the area under the curve y = 9 − x 2 over the interval [0,3] and n = 10, 20 and 50. Left end point approximation Right end point approximation Mid point approximation K xk * 9 − ( xk ) 2 xk * 9 − ( xk ) 2 xk * 9 − ( xk ) 2 1 0.0 9.000000 0.3 8.910000 0.15 8.977500 2 0.3 8.910000 0.6 8.640000 0.45 8.797500 3 0.6 8.640000 0.9 8.190000 0.75 8.437500 4 0.9 8.190000 1.2 7.560000 1.05 7.897500 5 1.2 7.560000 1.5 6.750000 1.35 7.177500 6 1.5 6.750000 1.8 5.760000 1.65 6.277500 7 1.8 5.760000 2.1 4.590000 1.95 5.197500 8 2.1 4.590000 2.4 3.240000 2.25 3.937500 9 2.4 3.240000 2.7 1.710000 2.55 2.497500 10 2.7 1.710000 3.0 0.00000 2.85 0.877500 n =16.605000 =18.022500 ∆x ∑ f ( xk ) = 19.30500 ∗ k =1 © Copyright Virtual University of Pakistan 189 28-Area as Limits VU And for different values of n we have the following approximations. Left end point Right end point Mid point n approximation approximation approximation 10 19.305000 16.605000 18.02250 20 18.663750 17.313750 18.005625 50 18.268200 17/728200 18.000900 © Copyright Virtual University of Pakistan 190 29- Definite Integral VU Lecture # 29 The Definite Integral Definition of Definite Integral Definite Integral of continuous functions with nonnegative values Definite Integral of continuous functions with negative and positive values Definite Integral of functions with discontinuities Properties of the Definite Integral Inequalities involving Definite Integral We have so far developed a definition of Area as limit of a bunch of rectangles with equal widths. The question arises that why choose widths? The answer is that we don’t have to. We want now to have a definition of Area that includes the general case where the widths of the rectangles are not necessarily equal. When the rectangles widths were equal note that the formula for the width was defined as b−a ∆x = n In this formula you can see that as n goes to infinity, the width delta x goes to zero If the widths are not the same for all rectangles, then this is not necessarily the case. Suppose that we have a rectangle construction in which we divide [a,b] so that one half is continually subdivided into smaller Rectangle, while the right is left as one big rectangle. In this case, when n goes to inf, left part goes to 0, not the Right half. Let's fix this problem. Subdivide [a,b] into n subinterval whose widths are ∆x1 , ∆x2 , ∆x3 ,..., ∆xn The subintervals are said to partition the interval, and the largest subinterval width is called the mesh size of the partition. This is denoted by max ∆xk © Copyright Virtual University of Pakistan 191  29- Definite Integral VU Read as “maximum of the ∆xk.” In this figure 9 5 max ∆xk =∆x3 = − =2 2 2 We see that If [a,b] is partitioned into n subintervals, and if xk is an arbitrary point in kth subinterval, then f ( xk * )∆xk is the area of the rectangle with height f( xk ) and width n delta xk , and ∑ f (x k =1 k * )∆xk is the sum of the shaded rectangular areas Now if we let max ∆xk → 0 the width of EVERY rectangle tends to 0 because none of them exceeds the max. So we have the area under the curve now defined more generally as n =A lim max ∆xk →0 ∑ f (x k =1 k * )∆xk DEFINITION 5.6.1 (Area under a curve) If the function f is continuous an [a, b] and if f ( x) ≥ 0 for all x in [a, b], then the area under n the curve y = f ( x) over the interval [a, = b] is defined by A lim max ∆xk →0 ∑ f ( x )∆x k =1 * k k Definite Integral of continuous functions with nonnegative values The limit in the definition we just saw is VERY important. It has special notation. We write it as © Copyright Virtual University of Pakistan 192 29- Definite Integral VU b n ∫ f ( x)dx = lim max ∆xk →0 ∑ f (x k =1 k * )∆xk a The expression on the left is called the definite integral of f from a to b. a and b are called the upper and lower limits of integration respectively. There is a relationship btw this definite integral and the indefinite integral we talked about earlier. We will see it later. So with this notation we can say that b ∫ f ( x)dx = Area under the curve y = f ( x) over [a, b]. a Our goal will be to evaluate the definite integral efficiently instead of using the definition all the time. Example 4 ∫ ( x − 1)dx 2 This represents the area under the curve y = (x – 1) over [2,4]. Here is a picture of it. Note that the region described here is just a trapezoid with h = 2, and b 1 = 1 and b 2 = 3. So we can evaluate this integral from 4 1 basic geometry as ∫ ( x − 1)= 2 dx 2 (2)(1 += 3) 4 n The sum ∑ f (x k =1 k * )∆xk is called the Riemann Sum in honor of the German mathematician Bernhard Riemann. The formula for the area we just saw was for continuous and nonnegative functions. It involves a limit. The question is: does the limit always exist, so that we can always find the area ? It is proved in advanced levels that for continuous and nonnegative functions, this limit always exists. Definite Integral of continuous functions with negative and positive values © Copyright Virtual University of Pakistan 193 29- Definite Integral VU Now we want to extend our area definition to include continuous functions on [a,b] that have both positive and negative values on [a,b]. Look at the following figure The rectangles below [a,b] but above the curve f(x) have some area just the way their counterparts above [a,b] but below curve f(x) do. The difference is that we can view the number representing the area of the rectangles below as negative values of f(x) for some xk*. So the Riemann sum gives 6 ∑ f (x k =1 k * )∆= xk f ( x1* )∆x1 + f ( x2* )∆x2 +... f ( x6* )∆x6 = A1 + A2 − A3 − A4 + A5 + A6 = ( A1 + A2 + A5 + A6 ) − ( A3 + A4 ) This says that the Riemann sum is the difference of two areas: the total area of rectangle above the x-axis, minus the total areas of rectangles below the x-axis. Now as max delta x goes to 0, the situation still works out nicely with the large # of rectangles filling in any left over space between the earlier rectangles and the curve. So we have the following definition Definition 5.6.2 If the function f is continuous on [ a,b], and can assume both positive and negative values ,then the net signed area A between y= f (x) and the interval [ a,b ] is defined by © Copyright Virtual University of Pakistan 194 29- Definite Integral VU b n ∫ f ( x)dx = lim max ∆xk →0 ∑ f (x k =1 k * )∆xk a Net signed area means that the total difference of the two areas above and below may be negative. If that happens then we just treat the negative as representing the fact that the area below is larger than the area above. EXAMPLE 4 ∫ (1 − x)dx 2 Geometrically, this region lies below the interval [2,4]. The region is a trapezoid and we can find the area as 4 using its dimensions. Since its below the x-axis we write 4 ∫ (1 − x)dx = 2 −4 Definite Integral of functions with discontinuities For the two definitions we have seen so far, f was a continuous function. This guaranteed that the are defined as a limits could always be found since the limits existed for continuous f. If f is not continuous, then the area may or may not be found as the limit of the Sum may or may not exist. Here is a definition for such a function for the area. DEFINITION 5.6.3 (Area under a curve) If the function f is defined on the close interval [a, b] then, f is called Riemann integrable on [a, b] or more simply integrable on [a, b] if the limit exists n lim max ∆xk →0 ∑ f ( x )∆x k =1 * k k If f is integrable on [a, b], then we define The definite integral of f from a to b by b n ∫ f ( x)dx = lim max ∆xk →0 ∑ f ( x )∆x k =1 * k k a © Copyright Virtual University of Pakistan 195 29- Definite Integral VU So far we assumed that the upper limit of integration was greater than the lower limit. Here is a definition that allows for the limits to be the same, and the case where the upper may be less than the lower. DEFINITION 5.6.4 (a) If a is in the domain of f , we define a ∫ f ( x)dx = 0 a If f is integrable on [a, b], then we define a b ∫ f ( x)dx = − ∫ f ( x)dx b a Properties of the definite Integral Here are some properties of the definite integral Theorem 5.6.5 If f and g are integrable on [a, b] and if c is a constant, then cf, f + g, and f – g are integrable on [a, b] and b b ∫ ∫ (a) cf ( x) dx = c f ( x) dx a a b b b ∫ (b) [ f ( x) + g ( x)]dx = a ∫ f ( x)dx + ∫ g ( x)dx a a b b b ∫ (c) [ f ( x) − g ( x)]dx = a ∫ f ( x)dx − ∫ g ( x)dx a a Now look at this figure It is clear from the picture that the area under the curve over [a,b] can be split into a sum of two areas, one area from a to c, and the other area from c to b. Formally © Copyright Virtual University of Pakistan 196 29- Definite Integral VU Theorem 5.6.5 If f is integrable on a closed interval containing the three points a, b and c then b c b ∫= a f ( x)dx ∫ f ( x)dx + ∫ f ( x)dx a c no matter how the points are ordered. Example 5 5 5 ∫ f ( x)dx = 1 −1, ∫ f ( x)dx = 3 3, ∫ g ( x)dx = 4 3 Find 3 ∫ f ( x)dx 1 From thm5.6.6 with a = 1, b = 5, c = 3 5 3 5 ∫= 1 f ( x)dx ∫ f ( x)dx + ∫ f ( x)dx 1 3 So 3 5 5 ∫ f ( x)dx =∫ f ( x)dx − ∫ f ( x)dx =−1 − 3 =−4 1 1 3 Inequalities involving definite integral Look at this figure This says that if a function is nonnegative (the graph does not go below x-axis) on [a,b], then the net area between the curve and x-axis must be greater than or equal to 0. Look at this figure © Copyright Virtual University of Pakistan 197 29- Definite Integral VU The graph of f does not go below that of g, or in other words f(x) >=g(x) over the interval and f and g are nonnegative, then area under f must be >= area under g over the interval Theorem 5.6.7 (a) If f is integrable on [ a,b ] and f (x) ≥ 0 for all x in [a,b ],then b ∫ f ( x) ≥ 0 a (b) If f and g are integrable functions on[ a,b] and f ( x) ≥ g ( x) for all x in b b [a, b] then ∫ f ( x) ≥ ∫ g ( x) a a Works with 0, but 2 x3 − 5 < 0. So f(x) as a whole is < 0. So from theorem 5.6.7a), with < instead of >= the integral is negative Definition 5.6.8 A function f is said to be bounded on interval [a, b] if there is a positive number M such that − M ≤ f ( x) ≤ M for all x in [a, b]. Geometrically, this means that the graph of f on the interval [a, b] lies between the lines y = -M and y = M. y = x 2 on the interval [-2, 2] is bounded since its graph lies between the lines y = 0 and y = 5 © Copyright Virtual University of Pakistan 198 29- Definite Integral VU Theorem 5.6.9 Let f be a function that is defined at all the points in the interval [ a,b]. (a) If f is continuous on [ a,b] ,then f is integrable on [ a,b ]. (b) If f is bounded on [ a,b ] and has only finite many points of discontinuity on [ a,b ] then f is integrable on [ a,b ]. (c ) If f is not bounded on [ a,b] ,then f is not integrable on [ a,b] © Copyright Virtual University of Pakistan 199 30- Fundamental Theorem of Calculus VU Lecture # 30 First Fundamental Theorem of Calculus 1st fundamental theorem of calculus. Relationship between definite and indefinite integrals. Mean Values theorem for Integrals. Average Values of a function. This is the 1st fundamental theorem of calculus. Theorem 5.7.1(The first fundamental theorem of calculus) If f is continuous on [a, b] and if F is an antiderivative of f on [a, b] , then b ∫ f ( x= a )dx F (b) − F (a ) This theorem tells us how to evaluate EASILY the definite integral. It says that find the anti-derivative of f(x)first, call it F(x), and then evaluate this function on the limits of the integration. Let's prove this. Proof We will use the Mean Value Theorem involving derivatives to prove the first fundamental theorem of calculus. b ∫ f ( x= a )dx F (b) − F (a ) Let us subdivide the given interval of integration namely [a, b] into n subintervals using the points x1 , x2 ,..., xn −1 in the interval [a, b] such that a < x1 < x2 <... < xn −1 < b. So now we have n subinterval of [a, b]. We can denote the widths of these intervals as ∆x1 , ∆x2 ,..., ∆xn. Where for example ∆x2 = x2 − x1. Since F ′( x) = f ( x) for all x in [a, b], it is obvious that F(x) satisfies the requirements of the Mean Value Theorem involving derivatives on each of the subintervals. So by MVT, we can find points x1* , x2* ,..., xn* In each of the respective subintervals [a, x1 ], [ x1 , x2 ] ,..., [ xn −1 , b ]. So we now make the following equations © Copyright Virtual University of Pakistan 200 30- Fundamental Theorem of Calculus VU ) F ′( x1* )( x1 − a= F ( x1 ) − F (a= ) f ( x1* )∆x1 F ( x2 ) − F ( x1 )= F ′( x2* )( x2 − x1 )= f ( x2* )∆x2 F ( x3 ) − F ( x2 )= F ′( x3* )( x3 − x2 )= f ( x3* )∆x3... F (b) − F ( xn −1 ) = F ′( xn* )(b − xn −1 ) = f ( xn* )∆xn Adding up these equations we get n F (b) − F (a= ) ∑ f ( x )∆x k =1 * k k Now increase n in such a way that max ∆xk → 0 Since f is assumed to be continuous, we have the following result n b = F (b) − F (a ) lim ∑ = f ( xk* )∆xk ∫ f ( x)dx max ∆xk → 0 k =1 a ↑ THEOREM 5.6.9(a) I did not apply the limits on the left because even if I did, the expression does not involve n and therefore nothing happens. F (b) − F (a ) can also be written as F ( x)]ba so we have b ∫ f ( x)dx = F ( x)] b a a Example 2 ∫ Evaluate xdx 1 2 ∫ xdx.The function F ( x) = 2 x 1 2 is and antiderivative of f ( x) = x.So we have 1 2 1 2  1 2 1 1 3 ∫1 xdx = 2 x 1 = 2 (4) − 2 (1) = 2 − 2 = 2 Here are a few properties of the bracket notation we just saw. Prove these yourself Properties [cF ( x)]a = c [ F ( x)]a b b [ F ( x) + G ( x)]a = [ F ( x)]a + [G ( x)]a b b b [ F ( x) − G ( x)]a = [ F ( x)]a − [G ( x)]a b b b © Copyright Virtual University of Pakistan 201 30- Fundamental Theorem of Calculus VU These are easy if you remember that this notation and the definite integral are the same thing!! Relationship between definite and indefinite integrals In applying the 1st theorem of Calc, it does not matter WHICH anti-derivative of f is used. If F is any anti-derivative of f then all the others have form F(x) + C by theorem 5.2.2 Hence we have the following: [ F ( x) + C ]=a [ F (b) + C ] − [ F (a) + C ] b b …(A) = F (b) − F (a ) = F ( x) ]a = ∫ f ( x)dx b a This shows that all anti-derivatives of f on [a,b] give the same values for the definite integral Now since ∫ f ( x= )dx F ( x) + C It follows from (A) that b b ∫ f ( x)dx =  ∫ f ( x)dx  a a So we can first evaluate the indefinite integral and use the limits as well to evaluate the definite integral This relates the definite and the indefinite integrals. Example Using the 1st theorem of calculus, find the area under the curve y = cos(x) over the interval [0, 2π] Since cos( x ) ≥ 0 for 0 ≤ x ≤ π , the area is π 2 π =A cos( x)dx  ∫ cos( x)dx  ∫= 2 0 0 π π = sin( x) ]02 = sin( ) − sin(0) = 1 2 I deliberately chose C = 0 since we just saw that the value of C=0 doesn’t change the answer. Example 3 ∫ (x − 4 x + 1)dx 3 Evaluate 0 © Copyright Virtual University of Pakistan 202 30- Fundamental Theorem of Calculus VU 3 3 ∫ ( x − 4 x + 1)dx=  ∫ ( x − 4 x + 1)dx  0 3 3 0 3 =  ∫ ( x3dx − ∫ 4 x.dx + ∫1.dx  0 3  x4 x2  81 21 =  − 4. + x  = ( − 18 + 3) − (0) = 4 2 0 4 4 Example 6  x2 x Work = Force x distance Distance unit is meter represented by m Force units are Pounds (lbs), Dynes or Newton (N). One Dyne = the force needed to give a mass of 1 gram an acceleration of 1 cm/s^2 1 Newton = the force needed to give a mass of 1 Kg an acceleration of 1 m/s^2 Example An object moves 5 ft along a line while subjected to a force of 100 lbs in its direction of motion. The work done is W = F.d = 100(5) = 500 ft.lbs. Work done by a variable force So far the force was a constant force. What if it’s changing constantly? The equation of Work we just saw will not work. We need Calculus © Copyright Virtual University of Pakistan 245 38-Work and Definite Integral VU In this figure, we have a block subjected to the force of a compressed spring. As the block moves from a to b, the spring gets un-compressed and the force it applied diminishes. So here is a case where the force varies with the position x of the spring. So the force F(x) is a function of the position of the block on the x-axis. We need to define work done by a variable force. This, in turn, will give us the answer to calculate it as well through definite integral. As before we subdivide the interval [a, b] into subintervals with coordinates a, x1 ,...xn −1 , b and widths ∆x1 etc Work and definite integral Let’s consider the kth interval and the force F over it. If the interval [ xk −1 , xk ] is small, then the force F will be almost constant on it. We can approx F on this interval by F(x k *) where x k * is a point in the kth interval. The width of this interval will be ∆xk © Copyright Virtual University of Pakistan 246 38-Work and Definite Integral VU So the work done on this interval is= W k F ( xk * )∆xk n n So the work done over the whole interval is= =k 1= Wk k 1 ∑ ∑ F (x k * )∆xk n b Let the largest subinterval go= to 0 to get W lim ∑ = F ( xk * )∆xk ∫ F ( x)dx max ∆xk → 0 k =1 a Example A cylindrical water tank of radius 10 ft and height 30 ft is half filled with water. How much force is needed to pump all the water over the upper rim of the tank? Solution Introduce a coordinate line as shown in figure, imagine the water to be divided into n thin layers with thicknesses ∆x1 , ∆x2 ,......., ∆xn How much force is required to move the Kth layer of water above the rim? © Copyright Virtual University of Pakistan 247 38-Work and Definite Integral VU The force required to move the Kth layer equals the weight of the layer, which can be found by multiplying its volume by weight density of water Pressure is defined as Force per unit Area. P = F / A = ph © Copyright Virtual University of Pakistan 248 38-Work and Definite Integral VU Where p is weight density and h is depth below surface of fluid. Fluid Pressure If a flat surface of area A is submerged horizontally in a fluid at a depth h, then the fluid exerts a force F perpendicular to the surface This is given by F = phA p is the weight density of the fluid. p for water is 62.4 lbs/ft^3 It is a fact from Physics that the shape of the container containing the fluid does not in any way effect F. If the three containers here have the same area for their bases and the fluid has the same height, then the F on the bases will be equal. Figure 6.8.1 Pascal’s Principle Fluid pressure is the same in all directions at a given height. Figure 6.8.2 © Copyright Virtual University of Pakistan 249 38-Work and Definite Integral VU If a flat surface is submerged horizontally, then the total force on its face can be measured easily since the pressure is the same at all the points since the height of these points does not vary. Suppose we submerge a flat surface in a fluid VERTICALLY. Then at each point along the height of the surface, the pressure will be different and therefore Force will be different. Here is where we need calculus Look at the picture We have a surface submerged in a tank. There is VERTICAL axis on the tank The surface is confined by x = a and x = b There is a height function of x that measures the depth of a point on the surface from the top of the tank. There is a width function of x that measures the width w (x ) of the surface at the height h (x). At different height (depth) Force on a section of the surface will be different. We want to know the total force on the flat surface. Subdivide the interval [a,b] into subintervals of various widths. © Copyright Virtual University of Pakistan 250 38-Work and Definite Integral VU Using the kth interval, which form a rectangle that will be used to approximate the force on that interval on the surface? The smaller the interval, the more accurate the approximation. The approximation will be made by Fk ≈ ph( xk * ) w( xk * )∆xk p = fluid density h( xk * ) = depth w( xk * )∆xk = area of rec tan gle F k is the force on the strip of width almost zero on the surface In the figure, imagine the rectangle to shrink to the dashed line. n n Hence, the total force in this fashion will be F = = Fk ≈ k 1=k 1 ∑ ∑ ph( xk * ) w( xk * )∆xk n b = Make the largest interval go to 0 to get F lim ∑ = ph( xk ) w( xk )∆xk * * ∫ ph( x)w( x)dx max ∆xk → 0 k =1 a Formula For Fluid Force Assume that a flat surface is immersed vertically in a liquid of weight density P and that the submerged portion extends from x=a to x=b on a vertical x-axis. For a1 uk # D $ * )  6. # 6 7 9; " 5 78( 0 ! ∞ 1 (−1) k +1 k =1 k " $ - # 1. # 1 )# 3 , # 1 6 7 9< " 5 78( 0 ! )! E # A ! * E 7 7 E ) # 6 7 9= " 5 78( 0 ! f d f ( x)dx = _____________ a b c d f ( x ) dx + f ( x ) dx + f ( x ) dx a b c b d f ( x ) dx + f ( x ) dx a c c d f ( x) dx + f ( x) dx a b d f ( x) dx a 6 7 9? " 5 78( 0 ! 1 f g b c [ f ( x) + g ( x)] dx = ______________ a  b b f (cx)dx + g (cx)dx a a b b f ( x) dx + g ( x)] dx a a b b c f ( x)dx + c g ( x)dx a a 6 79 " 5 78( 0 ! " b b f ( x)dx and f (t )dt a a $ 6 ! 6 ! ' $$ 6 ! F 6 7: " 5 78( 0 ! 2 2 f ( x) dx = 5 g ( x) dx = −3 −1 −1 2 [ f ( x) + 2 g ( x)] dx −1 $ 8 B == 6 7 :8 " 5 7 ( 1 1 1 1 + + +...... + 1 8 27 1000 3' # # 10 (1/ n3 ) n =1 6 7: " 5 7 ( / ! 7 ex 6 7 :9 " 5 7 ( 3 < 4 x dx 1 4 x dx 1 4 = x.1 dx 1 4 = x x + 1/ x.1 dx 1 6 7 :: " 5 79( 3 #< 6 (k 2 − 5) k =1 = −4 − 1 + 4 + 11 + 20 + 31 = 61 6 7 :; " 5 79( y = x2 , x > 0 ! y =1 y=4 ! ? 6 7 :< " 5 79( ∞ 3 an = 2 n n =5 6 #