Grade: The remainder of this page has been left blank for your workings. VERSION D. Midterm D: 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 D: Page of 2

2 Indefinite Integrals. 9 marks Each part is worth marks. Please write your answers in the boxes. a) Calculate the indefinite integral sinx) dx for < x < π/2. cosx) Answer: I = 2 cosx)+c Solution: Let u = cosx), so that sinx)dx = du. Then, I = u /2 du = 2 u+c. Using u = cosx) we get I = 2 cosx)+c. b) Calculate the indefinite integral x+ dx for x >. x 2 +x Answer: ln x ) + 2ln x+ ) +C Solution: The denominator x 2 + x factorizes as xx + ). Thus, the partial x+ fraction decomposition is xx+) = A x + B x+. Multiplying everything by xx+) we get x+ = Ax+)+Bx, and by plugging in the values x = and x = we obtain A = and B = 2. Then x+ x 2 +x dx = x + 2 x+) ) dx = ln x ) + 2ln x+ ) +C. Midterm D: Page 2 of 2

3 c) A Little Harder): Calculate the indefinite integral x 2 e x dx. Answer: e x x 2 2x 2)+C Solution: Let u = x 2 and dv/dx = e x. We calculate du/dx = 2x and v = e x, so that one step of integration by parts gives I = uv v du dx dx = x2 e x + 2xe x dx. Now we apply integration by parts again to J = 2xe x dx choosing u = 2x and dv/dx = e x, obtaining J = uv v du dx dx = 2xe x + 2e x dx = e x 2x 2)+C. Plugging this into our first equation for I we get I = x 2 e x +J = e x x 2 2x 2)+C. Midterm D: Page of 2

4 Definite Integrals 2. 2 marks Each part is worth 4 marks. Please write your answers in the boxes. a) Calculate π/2 cos x)dx. Answer: 2 Solution: Since the power of cosine is odd, we hold on to one copy of it and turn the rest into sines. We get π/2 cos x)dx = π/2 cosx) sin 2 x))dx. Now we can use the substitution u = sinx). We have du/dx = cosx), and that x = and x = π/2 map to u = and u =. This yields, π/2 b) Calculate cosx) sin 2 x))dx = 9x 2 x 2 +9 dx. Answer: 27 27π 4 x 2 u 2 )du = Solution: We first note that x 2 +9 = 9 x Then 9x 2 x 2 +9 dx = 9 9 ) dx = 9 8 x 2 +9 ] [u u = 2. x 2 +9 dx. To solve the second integral we use the substitution x = u, so that [ x 2 +9 = arctanu) 9u 2 +) du = Plugging this back into the first equation we get 9x 2 27π dx = 27 x ] = π 2. Midterm D: Page 4 of 2

5 c) A Little Harder): Calculate e 2 lnx x 2 dx. Answer: e 2 Solution: We use integration by parts, picking dv/dx = and u = lnx. We x2 compute v = x and du/dx =. Thus applying the IBP formula we get x I = uv v du [ dx dx = lnx ] e 2 + x x dx = 2 [ 2 e + ] e 2 = 2 x e 2. Midterm D: Page 5 of 2

6 Riemann Sum, FTC, and Volumes. 2 marks Each part is worth 4 marks. Please write your answers in the boxes. a) Calculate the infinite sum lim n n i= i 2 + i n n by first writing it as a definite integral. Then, evaluate this integral. ) Answer: Solution: We try to pick x = n,a =,b =,x i = i, so we have to write ) n i the summand in the form xf. By collecting a in the expression we get n n ) 2 ) i i + so we have fx) = x 2 +x n n n, and thus lim n n i= i 2 + i n n = lim n n xfx i ) = i= Using the substitution u = +x we get x 2 +x dx = 2 2 [ ] udu = u /2 2 = 2 x 2 +x dx. 2 ) 2. b) For x define Fx) and gx) by Fx) = x cos2 t)dt and gx) = xfx 2 ). Calculate g π). Answer: 5 2 π Solution: By the product rule, chain rule, and FTC I we have g x) = Fx 2 )+2x 2 F x 2 ) = Fx 2 )+2x 2 cos 2 x 2 ). Setting x = π, we get g π) = Fπ)+2πcos 2 π) = Fπ)+2π. Now, π π Fπ) = cos 2 +cos2t) t)dt = dt = [ t+ sin2t) ] π = π So we conclude that g π) = π 2 +2π = 5 2 π. Midterm D: Page 6 of 2

7 c) Write a definite integral, with specified limits of integration, for the volume obtained byrevolvingtheboundedregionbetweenx = y 2 andx = 4+y 2 aboutthevertical line x = 2. Do not evaluate the integral. Solution: The plot is as shown: Answer: V = π y2 ) 2 2+y 2 ) 2 )dy. The red curve is x = y 2, the blue curve is x = 4+y 2 and the green line is the axis of rotation x = 2. The red and blue curves meet where y 2 = 4+y 2, which gives us y = ± 2 and x = 2. A slice of the rotational solid at height y will be a circular crown with inner radius r y = 2+y 2 the distance between x = 2 and x = y 2 ) and outer radius R y = 6 y 2 the distance between x = 2 and x = 4+y 2 ). Thus the area of the slice will be A y = πr 2 y r 2 y) = π6 y 2 ) 2 2+y 2 ) 2 ). This gives the volume V = π y2 ) 2 2+y 2 ) 2 )dy. Midterm D: Page 7 of 2

8 4. a) 2 marks Plot the finite area enclosed by y 2 = x and x = 8 2y. Solution: The area is the region enclosed between the blue and red curves: b) 4 marks Write a definite integral with specific limits of integration that determines this area. Do not evaluate the integral. Answer: A = y y2 ) dy Solution: Thetwocurvesmeetwheny 2 = 8 2y, whichhassolutionsy = 2, 4. Seeingashowforbothcurvesxisexpressedasafunctionofy, wechoosetowrite the area as an integral in the variable y, with x B y) = y 2 and x T y) = 8 2y. By evaluating at y = we see that x T y) x B y) on the interval [ 4,2]. Then we get the integral 2 4 [x T y) x B y)] dy = 2 Alternatively, as an integral in x, the area is A = 2 4 xdx y y 2 ) dy. 4 x ) 2 + x dx. Midterm D: Page 8 of 2

9 5. A solid has as its base the region in the xy-plane between y = x 2 /9 and the x- axis. The cross-sections of the solid perpendicular to the x-axis are semi-circles with the diameter of the semi-circle in the base. a) 4 marks Write a definite integral that determines the volume of the solid. Answer: V = π 8 x2 /9) 2 dx Solution: The points of intersection with the x-axis are given by x 2 /9 =, so we get x = ±. A vertical slice of the solid at x will be a half circle whose diameter is x 2 /9, and thus will have area A x = π 2 2 x2 /9) ) 2. Then the volume is given by V = A x dx = π 8 x2 /9) 2 dx. b) 2 marks Evaluate the integral to find the volume of the solid. Answer: 2π 5 Solution: We simplify by using the substitution u = x/ so that V = π 8 x2 /9) 2 dx = π 8 u 2 ) 2 du. Now we note that the function is even, so that V = π 8 u 2 ) 2 du = π 4 u 2 ) 2 du. Finally we expand the formula and compute the integral: V = π 4 We calculate that u 2 ) 2 du = π 4 V = π 4 u 4 2u 2 + ) du = π ) + = 2 5 π. [ u 5 5 2u +u ]. Midterm D: Page 9 of 2