Grade: The remainder of this page has been left blank for your workings. VERSION E. Midterm E: Page 1 of 12

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1 First Name: Student-No: Last Name: Section: Grade: The remainder of this page has been left blank for your workings. Midterm E: Page of

2 Indefinite Integrals. 9 marks Each part is worth 3 marks. Please write your answers in the boxes. (a) Calculate the indefinite integral e x +e x dx. Answer: I = 3 (+e x ) 3/ +C Method I: Let u = +e x, so that e x dx = du. Then, udu I = = 3 u3/ +C. Using u = +e x we get I = 3 (+e x ) 3/ +C. Method II: Let u = +e x, so that u = +e x. Differentiating implicitly gives e x dx = udu. Then, I = u du = 3 u3 +C. Using u = +e x we get I = 3 (+e x ) 3/ +C. (b) Calculate the indefinite integral (x+)e x dx for x >. Answer: I = (x+)e x +C. Method I: Let u = (x + ) and dv/dx = e x. We calculate du/dx = and v = e x, so that one step of integration by parts gives I = uv v du dx dx = (x+)e x + e x dx = (x+)e x e x +C. Method II: First, write (x+)e x dx = xe x dx+ e x dx = xe x dx e x. Let u = e x, so that e x dx = du and logu = x. Then, xe x dx = log(u)du = ulog(u) u+c. Using u = e x we get xe x dx = xe x e x +C and hence (x+)e x dx = (x+)e x +C. Midterm E: Page of

3 (c) (A Little Harder): Calculate the indefinite integral tan 5 (x)sec 3 (x)dx. Answer: I = sec7 (x) 7 sec5 (x) 5 + sec3 (x) 3 +C. Method I: Let u = sec x, so that du = tan(x) sec(x) dx. Then, using the identity tan (x) = sec (x) = u, (u I = ) (u u du = 6 u 4 +u ) du = u7 7 u5 5 + u3 3 +C. Using u = sec(x) we get I = sec7 (x) sec5 (x) + sec3 (x) +C Method II: First note that I = sin 5 (x) cos 8 (x) dx. Let u = cosx, so that du = sin(x)dx. Then, using the identity sin (x) = cos (x) = u, ( u ) ( I = du = u 8 u 8 u + ) du = 6 u 4 7u 7 5u + 5 3u +C. 3 Using = sec(x) we get I = sec7 (x) sec5 (x) + sec3 (x) +C. u Method III: Let u = sec (x), so that sec (x)tan(x)dx = du. We assume that sec(x) > so that u = sec(x). The case of sec(x) < follows by plugging u = sec(x). Then, using the identity tan (x) = sec (x) = u, we get I = (u ) udu = (u 5/ u 3/ +u /) du = u7/ 7 u5/ 5 +u3/ 3 +C. Using u = sec(x) we get I = sec7 (x) 7 sec5 (x) 5 + sec3 (x) 3 +C. Midterm E: Page 3 of

4 Definite Integrals. marks Each part is worth 4 marks. Please write your answers in the boxes. (a) Calculate I = π/8 sin (x)dx. Answer: I = π 6 8. Use sin (x) = cos(x) to get I = π/8 cos(4x) [ x dx = sin(4x) 8 π/8 x= = π 6 8. (b) Calculate I = e x lnxdx. Answer: I = e Let u = lnx and dv/dx = x. We calculate du/dx = x so that one step of integration by parts gives I = x3 3 lnx e x= e x 3 and v = x3 3, e3 dx = 3 x3 9 e x= = e3 3 e = e Midterm E: Page 4 of

5 (c) (A Little Harder): Calculate I = e x sin(x)dx. We first recall that I = Answer: I =. e x sin(x)dx = lim L L e x sin(x)dx. We compute the indefinite integral e x sin(x)dx. Let u = sin(x) and dv/dx = e x. We calculate du/dx = cos(x) and v = e x, so that one step of integration by parts gives e x sin(x)dx = e x sin(x)+ e x cos(x)dx. Let u = cos(x) and dv/dx = e x. We calculate du/dx = sin(x) and v = e x, so that another step of integration by parts gives e x sin(x)dx = e x sin(x) e x cos(x) e x sin(x)dx Moving e x sin(x)dx to the left-hand-side and dividing by yields e x sin(x)dx = e x (sin(x)+cos(x)). It follows that [ ] [ ] e x I = lim L (sin(x)+cos(x)) L x= = lim L e L (sin(l)+cos(l) =. Midterm E: Page 5 of

6 Riemann Sum, FTC, and Volumes 3. marks Each part is worth 4 marks. Please write your answers in the boxes. (a) Calculate the infinite sum lim n n i= 8i /n e 4i n by first writing it as a definite integral. Then, evaluate this integral. Answer: e 4 We identify a =, b =, x = /n, x i = i/n, and f(x i ) = 8x i e 4x i. This yields S lim n n i= 8i /n = lim e 4i n n n ( x)f(x i ) = i= 8xe 4x dx. To calculate the integral we let u = 4x, which yields S = e 4x x= = e 4. (b) For x > define F(x) = x t / dt and g(x) = F(x ). Calculate g (). Answer: g () =. We use the product rule to get g (x) = F(x ) F (x )x. Now, by FTC I, we get F (x ) = (x ) / =. This yields, x g (x) = F(x ). () Now, let x = and calculate that F(4) = Therefore, from (), we get 4 [ t / dt = t 4 = 4 =. t= g () = F(4) =. Midterm E: Page 6 of

7 (c) Write a definite integral, with specified limits of integration, for the volume obtained by revolving the bounded region between x = (y ) and x = (y ) about the vertical line x =. Do not evaluate the integral. Answer: V = π 3 [ (y ) ] dy The two curves intersect when (y ) = (y ), which yields (y ) = so that y = ±. This gives y = and y = 3 for both x =. The intersection points are in the first quadrant. Define x R = (y ) (right blue curve) and x L = (y ) (left red curve). Then, at each y in [,3], we have that (x R +) and (x L +) are the distances of the two curves from the axis of rotation x = shown by the orange curve. This yields V = π = π 3 3 = π 3 [ (xr +) (x L +) ] dy, [ (4 (y ) ) ( (y ) + ) ] dy, [ (y ) ] dy. Midterm E: Page 7 of

8 4. (a) marks Plot the finite area enclosed by y = 4 x and x = 3y 6. The area is the region enclosed by the two curves in the plot: (b) 4 marks Write a definite integral with specific limits of integration that determines this area. Do not evaluate the integral. Answer: 5 [(4 y ) (3y 6)] dy. To find the intersection points we set y = 4 x = 4 (3y 6). This yields, y +3y = (y )(y +5) =, which gives y = 5 and y =. We label x T = 4 y (red curve) and x B = 3y 6 (blue curve), and observe that x T > x B on 5 < y <. The area is best calculated as an integral in y, so that A = (x T x B ) dy = 5 5 [ (4 y ) (3y 6) ] dy. Alternatively, the area can be calculated as an integral in x, so that ( x+6 A = ) 4 4 x dx+ 4 xdx. 3 Midterm E: Page 8 of

9 5. A solid has as its base the region in the xy-plane between y = x /36 and the x-axis. The cross-sections of the solid perpendicular to the x-axis are squares. (a) 4 marks Write a definite integral that determines the volume of the solid. The intersection points with the x-axis are x = ±6. This gives, V = 6 A(x)dx as the volume, where A(x) is the cross-sectional area of the 6 solid at position x. This cross-section is a square that has area A(x) = [y(x)]. Here we have used the fact that the area of a square with side of length b is b. This gives, 6 6 ] V = [y(x)] dx = [ x dx (b) marks Evaluate the integral to find the volume of the solid. Answer: 3/5 Since the integrand is even, we write V = [ 6 x 36] dx. Now put x = 6u, so that dx = 6du, and so ( V = ( u ) du = u +u 4) ( du = 3 + ) = Midterm E: Page 9 of