MMCAS Unit 2 Practice Test Integration Solutions PDF

# Practice Test Mathematical Methods CAS Unit 2 ## Anti-differentiation test - Technology Free **Name:** _Solutions_ **Total Marks:** /22 **Instructions:** - Writing time: 20 minutes. - Technology free; use of a CAS calculator is not allowed. - All relevant working must be shown and clearly se...

# Practice Test Mathematical Methods CAS Unit 2 ## Anti-differentiation test - Technology Free **Name:** _Solutions_ **Total Marks:** /22 **Instructions:** - Writing time: 20 minutes. - Technology free; use of a CAS calculator is not allowed. - All relevant working must be shown and clearly set out in the spaces provided. ## Question 1 Find the integral for each of the following with respect to x: a. $\int 6x^2 + 4x dx$ = $\frac{6x^3}{3} + \frac{4x^2}{2} + C$ = $x^3 + 2x^2 + C$ b. $\int 2(x+2)^2 dx$ = $\int 2(x^2 + 4x + 4) dx$ = $2 (\frac{x^3}{3} + \frac{4x^2}{2} + 4x) + C$ = $\frac{2}{3}x^3 + 2x^2 + 8x + C$ c. $\int \frac{x^3 + 3}{x^2} dx$ = $\int (\frac{x^3}{x^2} + \frac{3}{x^2}) dx$ = $\int (x + 3x^{-2}) dx$ = $\frac{x^2}{2} + 3\frac{x^{-1}}{-1} + C$ = $\frac{x^2}{2} + \frac{3}{x} + C$ (1+2+2 = 5 marks) ## Question 2 Find f(x) given f'(x) = 6x - 3 and f(1) = 2. f(x) = $\frac{6x^2}{2} - 3x + C$ f(x) = $3x^2 - 3x + C$ f(1) = 3-3+C = 2 C = 2. ::f(x) = $3x^2 - 3x + 2$ (3 marks) ## Question 3 a. Sketch the graph $y = x^2 + 2$ A graph of a parabola opening upward with a vertex at (0,2) passing through point (2,6). b. Hence, find the area under the curve from x = 0 to x = 2. A = $\int_0^2 (x^2+ 2) dx$. = $[\frac{x^3}{3} + 2x]_0^2$ = $(\frac{8}{3} + 4)-0$ A = $\frac{20}{3}$ sq. uni5 (2 + 3 = 5 marks) ## Question 4 A curve is defined by the function f(x) = 9 - x² and has a domain of [0,3]. f(x) is shown below; A graph of an upside down parabola crossing the y-axis at y=9 with a vertex at (0,9) and x-intercepts at x = 3 and x = -3. An approximation to the area beneath the curve is to be made using the Trapezoidal estimate and three equally spaced intervals. a. Draw the appropriate trapeziums on the diagram above. Three trapeziums, with bases along the x-axis, starting at 0, and ending at 3, are drawn under the graph. b. Evaluate the estimated area enclosed by f(x) and the x-axis. A = $\frac{3-0}{2x3}$ x (9+2(8+5)+0) = $\frac{1}{2}$ (9+26) = $\frac{35}{2}$ sq units (1 + 2 = 3 marks) ## Question 5 Consider the function f(x) = $1-x^3$ a. Sketch the graph f(x), and shade the area bound by the x-axis, x = -1 and x = 2 A graph of a cubic function whose y-intercept is (0,1). The graph crosses the x-axis at the point (1,0) and goes below the x-axis. The area is shaded between x=-1 and x=2. b. State the integral expression which represents this area. A= $\int_{-1}^1 f(x) dx - \int_1^2 f(x) dx$. c. Find the area from your expression found in b. A = $[\frac{x^4}{4} - x]_0^1 - [\frac{x^4}{4} -x]_1^2 = [1 -\frac{1}{4} - 1] - [2-4-1 +4]$ = $1-\frac{1}{4} + 1 - 4 - (2-4-1+4)$ = $5-\frac{1}{4}$ = $\frac{19}{4}$ sq. uni5 (2 + 1 + 3 = 6 marks) # End of technology free test # Practice Test Mathematical Methods CAS Unit 2 ## Antidifferentiation test - Technology Active **Name:** **Total Marks:** /24 **Instructions:** - Writing time: 25 minutes. - Technology active; use of a CAS calculator is allowed. - One bound book of notes is allowed. - All relevant working must be shown and clearly set out in the spaces provided. ## Part A: Multiple Choice (Circle one alternative, one mark each). 1. $\int 3x^2 - 3x + 1 dx$ is A. 6x - 3 B. $x^3 - 3x^2 + x + C$ C. 2x + 1 D. $x^3 - 3$ **E. $x^3 - 3x^2 + x + C$** 2. If $y = 2x + 4$ and $y = 1$ when $x = 0$, an expression for y is A. $y = x^2 + 4x$ **B. $y = x^2 + 4x + 1$** C. $y = 2$ D. $y = x^2 + 4x - 1$ E. $y = 2x^2 + 4x - 5$ 3. If the curve with equation $y = f(x)$ for which $f'(x) = 6x + k$ has a stationary point at (1, 2), $f(x)$ equals A. $3x^2 - 6x$ B. $3x^2 + x$ **C. $3x^2 - 6x + 5$** D. $3x^2 + 6x - 4$ E. $3x^2 - 6x + 4$ 4. $\int_0^a (2x + 4) dx = 12$, where a is a positive real number. The value of a is A. 4 B. 3 C. $\frac{11}{2}$ D. 6 **E. 2** 5. $\int_{-1}^2 (3f(x) -1) dx$ can be written as **A. $3 \int_{-1}^2 f(x) dx - 1$** B. $3 \int_{-1}^2 (f(x)-1) dx$ C. $3 \int_{-1}^2 f(x) dx -\int_{-1}^2 1 dx$ D. $3\int_{-1}^2 f(x) dx - 3$ E. $3\int_{-1}^2 f(x) dx + 3$ 6. The area of the shaded region shown is A graph of a curve and a line. The shaded area is above the line and below the curve. A. $\frac{5}{3}$ B. -4 **C. $\frac{9}{2}$** D. $-\frac{2}{3}$ E. 3 7. The graph with the equation $y = k(x −3)²$ is shown. If the area of the shaded region is 36 square units, the value of k is A graph of a parabola opening upward with a vertex at (3,0) and a shaded area under the curve. **A. 4** B. -4 C. 9 D. $\frac{4}{7}$ E. 32 8. The graph with equation $y = f(x)$ is shown. The total area of the shaded region is equal to A graph of a curve crossing the x-axis at x=1, x=3, x=4. The shaded area is between x=1 and x=3. A. $\int_1^3 f(x)dx$ **B. $\int_1^3 f(x)dx - \int_3^4 f(x) dx.$** C. $-\int_0^1 f(x)dx + \int_1^3 f(x)dx$ D. $\int_0^3 f(x)dx - \int_3^4 f(x) dx$ E. $-\int_0^1 f(x)dx + \int_1^4 f(x)dx$ ## Part B: Extended response (Show full working where required). ## Question 1 (9 marks) 1. An area of agricultural land is 10 by 10 kilometres and is crossed by a river from East to West. The river splits the land into two regions and is modelled by the function; $r (x) = \frac{1}{10} (x-3)² +8$ The region of land to the south of the river (shaded) is used for farming but the region to the north of the river is forested. The land is shown in the diagram below. A graph of a parabola opening upward with a vertex at (3,8). The shaded area is below the curve. The area is bounded by the x-axis. For a farm to be economically viable the farmer needs the ratio of productive farmland to non- productive farmland to be greater than 2:1. a. By drawing 5 trapeziums on the diagram, calculate an approximate value for the area of farmland using the Trapezoidal Rule. A = $\frac{10-0}{2 \times 5} \times (\frac{1}{10} + 2 (\frac{31}{10} + \frac{39}{10} + \frac{31}{10} + \frac{1}{10})) = 67 sq. km.$ b. Would the sum of these Trapeziums give an under or over-estimate of the area? Explain. Underestimate as the trapeziums are inside the curve. c. Calculate the exact value of the area of productive farmland using calculus. A = $\int_0^{10} (\frac{1}{10}(x-3)^2 + 8) dx = [\frac{1}{30}(x-3)^3+ 8x]_0^{10}$ = $(\frac{343}{30}+80) - (-\frac{27}{30} + 0)$ = $\frac{203}{3}$ km² d. Is the farm likely to be economically productive? Give a reason. $\frac{203}{3} : 10 \approx 2.09:1$ The ratio is greater than 2:1, farm is economically viable. ## Question 2 (7 marks) A stadium has a parabolic shaped roof with given function f: [-25,25] → R, f(x) = 20-$\frac{3}{125}$x². The diagram of the stadium is shown below. A graph of an upside down parabola with the vertex at the point (0,20) crossing the x-axis at x=25. The stadium is 75 metres in length. a. Write down the coordinates of point E. (0,20) b. Write down the coordinates of points C. (25,5) c. Using calculus, find the cross-sectional area ABCDE of the stadium to nearest square metres. A = $\int_{-25}^{25} (20-\frac{3}{125}x^2) dx$ = $[20x-\frac{3x^3}{125\times3}]_{-25}^{25}$ = $500-125 - (-500+125)$ = $500-125+500-125$ = $1000-250$ = $750$ sq. metres. d. Hence, find the volume of the stadium. V = $A\times h = 75 \times 750 = 56250 m^3$ e. The stadium is to have several air-conditioners strategically placed around it. Each one can service a volume of 11, 250 m³. How many air-conditioners are required? No. of airconditioners = $\frac{56250}{11250}$ = 5. ## End of technology active test