Final Exam Review Sheet - Calculus II for Business & Social Sciences | MATH 242, Study notes of Mathematics

Download Final Exam Review Sheet - Calculus II for Business & Social Sciences | MATH 242 and more Study notes Mathematics in PDF only on Docsity! Business Calculus II Final Review Sheet Remember our final is on TUESDAY MARCH 15 at 3.15pm. 1. Calculate the following indefinite integrals. (a) ∫ (2x + 1 + x−1)dx. (b) ∫ 3x2+2 x3+2xdx. (c) ∫ 3x2+2 (x3+2x)3 dx. (d) ∫ (3x2 + 2)ex 3+2xdx. (e) ∫ 3x2+2 x3+2x ln(x 3 + 2x)dx. Solution. (a) x2 + x + ln |x| + c. (b) Let u = x3 + 2x. Then dudx = 3x 2 + 2. So (3x2 + 2)dx = du. Hence the integral transforms to R du u = ln |u| + c = ln |x3 + 2x| + c. (c) Let u = x3 + 2x again, so (3x2 + 2)dx = du. The integral transforms toR du u3 = − 1 2u2 + c = − 1 2(x3+2x)2 + c. (d) Let u = x3 + 2x again, so (3x2 + 2)dx = du. The integral transforms toR eudu = eu + c = ex 3+2x + c. (e) Let u = ln(x3 + 2x), so dudx = 3x2+2 x3+2x , so du = 3x 2+2 x3+2x dx. The integral transforms to R udu = u 2 2 + c = (ln(x3+2x))2 2 + c. 2. Calculate the following definite integrals. (a) ∫ 1 0 x 3dx. (b) ∫ 1 0 x2 x3+1dx. (c) ∫ 1 0 xe x2dx. Solution. (a) [x 4 4 ] 1 0 = 1 4 . (b) Let u = x3 + 1. Then du = 3x2dx. So x2dx = 13du. The integral transforms to 13 R 2 1 du u = 1 3 [ln u] 2 1 = 1 3 ln 2. (c) Let u = x2. Then du = 2xdx, so xdx = 12du. The integral transforms to 1 2 R 1 0 eudu = 12 [e u]10 = 1 2 (e− 1). 3. Calculate the areas of the following regions. Do not count areas below the x-axis as negative! (a) The region enclosed by y = x and y = −x4. (b) The region between y = x, y = x3, x = −1 and x = 1. (c) The region enclose by y = ex, y = 2 and the y-axis. Solution. 1 (a) Where do they cross? x = −x4 so x = 0, y = 0 or x = −1, y = −1. So we need to compute R 0 −1(−x4 − x)dx. (You should sketch the graph to see what is going on.) This is [−x5/5−x2/2]0−1 = −(1/5−1/2) = 1/2−1/5 = 5/10−2/10 = 3/10. (b) Where do y = x and y = x3 cross? x = x3 if x = 0 or x = 1 or x = −1. If you draw the picture you see that this region is made of two equal areas. So the total area is twice the right hand area, i.e. 2 R 1 0 (x− x3)dx = 2[x2/2− x4/4]10 = 2(1/2− 1/4) = 1/2. (c) Again you need to draw a sketch. If you do you see that you can compute this area by subtracting R ln 2 0 exdx from the area of the rectangle with base ln 2 and height 2. So the area is 2 ln 2− [ex]ln 20 = 2 ln 2− 2 + 1 = 2 ln 2− 1. 4. Sales of ice cream are continuously rising at a rate of 10 percent per month. My ice cream company currently sells 1000 quarts each month. Write down a differential equation describing the change in sales and then solve it to predict my monthly sales in 6 months time. Solution. Let s(t) be monthly sales after t months. The initial condition is that s(0) = 1000. The differential equation is s′(t) = 0.1s(t). So dsdt = 0.1s. Separate variables to get R ds s = 0.1 R dt, so ln s = 0.1t + c. Apply e? to both sides to get s = Ae0.1t for some new constant A (= ec). Plug in t = 0 to get 1000 = A. Hence s = 1000e0.1t. So in 6 months time the monthly sales will be s = 1000e0.6 = 1822 quarts. 5. Find the equation for the tangent plane to the following surfaces at the point (1, 2, 3). (a) z = x2 + 3xy − 2y − 3. (b) z = 7x− 3y + 2. Solution. (a) ∂z∂x = 2x + 3y, ∂z ∂y = 3x + 2 y2 . At x = 1 and y = 2 these give 8 and 3.5. So the tangent plane is z = 8x + 3.5y+?. To find the ? put x = 1, y = 2, z = 3 to get 3 = 15+? so ? = −12. So its z = 8x + 3.5y − 12. (b) Duh. Its already a plane. So there is nothing to do. The answer is z = 7x− 3y + 2. 6. Find all the critical points of the following function, and decide if they are local minima, local maxima or neither. (a) f(x, y) = x2y − 2x2 − 4y2. (b) f(x, y) = x2 + y − ey. (c) f(x, y) = x2 + y2 + 2xy . (d) f(x, y) = ex 2+y2 . Solution. (a) fx = 2xy − 4x = 0, fy = x2 − 8y = 0. We need to solve these simultaneous equations. From the first one, we have x(2y − 4) = 0 so either 2