Calculus Past Paper PDF

## Calculus (3) ### Double Integrals #### **Visual Representation** - **Horizontal strip:** - The double integral is represented by the area under the curve of the function *f(x,y)*. - The integration is done first with respect to *x* and then with respect to *y*. - The limits of int...

## Calculus (3) ### Double Integrals #### **Visual Representation** - **Horizontal strip:** - The double integral is represented by the area under the curve of the function *f(x,y)*. - The integration is done first with respect to *x* and then with respect to *y*. - The limits of integration for *x* are from *x = f1(y)* to *x = f2(y)*. - The limits of integration for *y* are from *y = a* to *y = b*. - **Vertical strip:** - The double integral is represented by the area under the curve of the function *f(x,y)*. - The integration is done first with respect to *y* and then with respect to *x*. - The limits of integration for *y* are from *y = f1(x)* to *y = f2(x)*. - The limits of integration for *x* are from *x = a* to *x = b*. ### Example: Find the area bounded by y²=2x & y=x **1. Sketch:** **2. Points of intersection:** - Solve the simultaneous equations y²=2x and y=x. - Substituting y=x into y²=2x, we get x²=2x - This simplifies to x²-2x=0 - Factoring out x, we get x(x-2)=0 - Therefore, the solutions are x=0 or x=2. - This gives us the points of intersection (0,0) and (2,2). **3. Double integrals** **Horizontal strip:** - I = ∫∫(1)dx dy - The limits of integration for *x* are from *x=y²⁄2* to *x=y*. - The limits of integration for *y* are from *y=0* to *y=2*. - Therefore, I = ∫(from y=0 to y=2) ∫(from x=y²⁄2 to x=y) (1) dx dy - Solving the inner integral first: - ∫(from x=y²⁄2 to x=y) (1) dx - = [x] (from x=y²⁄2 to x=y) - = y - y²/2 - Now solve the outer integral: - ∫(from y=0 to y=2) (y - y²/2) dy = [y²/2 - y³/6] (from y=0 to y=2) = 2/3 **Vertical strip:** - I = ∫∫(1))dy dx - The limits of integration for *y* are from *y=x* to *y=√2x*. - The limits of integration for *x* are from *x=0* to *x=2*. - Therefore, I = ∫(from x=0 to x=2) ∫(from y=x to y=√2x) (1) dy dx - Solving the inner integral first: - ∫(from y=x to y=√2x) (1) dy - = [y] (from y=x to y=√2x) - = √2x - x - Now solve the outer integral: - ∫(from x=0 to x=2) (√2x - x) dx = [(2/3)√2x³/2 - x²/2] (from x=0 to x=2) = 2/3 ### Example (2): Find the area bounded by y=√x & y = x - 2 **1. Sketch:** **2. Points of intersection:** - Solve the simultaneous equations y=√x and y=x-2. - Substituting y=√x into y=x-2, we get √x= x-2 - Squaring both sides, we get x = x² - 4x + 4 - This simplifies to x² - 5x + 4 = 0 - Factoring out the equation, we get (x-4)(x-1) = 0 - Therefore, the solutions are x=1 or x=4. - This gives us the points of intersection (1,1) and (4,2). **3. Double integrals** **Vertical strip:** - The area will be split into two sections to be solved individually, as the lower limit of the first section is y=-√x (from x=0 to x=1), and the upper limit of the second section is y=√x (from x=1 to x=4). - I = ∫(from x=0 to x=1) ∫(from y=-√x to y=√x) (1) dy dx + ∫(from x=1 to x=4) ∫(from y=x-2 to y=√x) (1) dy dx - Solving each integral individually. - Part 1: = [y] (from y=-√x to y=√x) = 2√x (from x=0 to x=1) = 2 - Part 2: = [y] (from y=x-2 to y=√x) = √x - (x-2) (from x=1 to x=4) = 9/2 - Therefore, I = 2 + 9/2 = 13/2 **Horizontal strip:** - I = ∫∫(1) dx dy - The limits of integration for *x* are from *x=y²* to *x= y+2*. - The limits of integration for *y* are from *y=-1* to *y=2*. - Therefore, I = ∫ (from y=-1 to y=2) ∫ (from x=y² to x=y+2) (1) dx dy - Solving the inner integral first: - ∫ (from x=y² to x=y+2) (1) dx - = [x] (from x=y² to x= y+2) - = y+2 - y² - Now solve the outer integral: - ∫ (from y=-1 to y=2) (y+2 - y²) dy = [y²/2 + 2y - y³/3] (from y=-1 to y=2) = 9/2 ### Change the order of integration - Changing the order of integration is done by changing the limits of integration for *x* and *y*, while keeping the same function. - This is useful if the original order of integration makes the integral difficult to solve. ### Example: Evaluate ∫(from y=0 to y=2x) ∫(from x=0 to x=1) e^x dy dx **1. Change the order of integration** - The original order of integration is with respect to *y* first and then *x*. We need to change it to with respect to *x* first and then *y*. - The new limits of integration for *x* will be from *x = 0* to *x = y/2*. - The new limits of integration for *y* will be from *y = 0* to *y = 2*. - Therefore, the new integral is: ∫(from y=0 to y=2) ∫(from x=0 to x=y/2) e^x dx dy **2. Solve the integral** - I = ∫(from y=0 to y=2) ∫(from x=0 to x=y/2) e^x dx dy - Solving the inner integral first: - ∫(from x=0 to x=y/2) e^x dx - = [e^x] (from x=0 to x=y/2) - = e^(y/2) - 1 - Now solve the outer integral: - ∫(from y=0 to y=2) (e^(y/2)-1) dy = [2e^(y/2) - y] (from y=0 to y=2) = 2e - 2