MATH 1014 3.00 PW – APPLIED CALCULUS II – Prof. Madras Solutions for Practice Test #1 1. (a) Integrate by parts: Let u = ln x and dv = x−1/2 dx. Then du = (1/x) dx and v = 2x1/2 , and we get Z Z ln x 1 1/2 √ dx = 2x ln x − 2x1/2 dx x x = 2x1/2 ln x − 4x1/2 + C . (b) Use partial fractions. Write 3x + 4 (x − 2)(x + 3) = = A B + x−2 x+3 Ax + 3A + Bx − 2B (x − 2)(x + 3) Matching coefficients gives A + B = 3 and 3A − 2B = 4. The solution of this pair of equations is A = 2 and B = 1. So Z Z 3x + 4 2 1 dx = + dx = 2 ln |x − 2| + ln |x + 3| + C. (x − 2)(x + 3) x−2 x+3 √ √ 2. Let x = tan t. Then dx = sec2 t dt, and 1 + x2 = 1 + tan2 t = sec t. Thus Z Z Z sec t 1 1 2 √ sec t dt = dx = 2 2 dt . 2 2 tan t sec t tan t x 1+x It is not clear what to do with this, so try writing it in terms of sin and cos: Z Z 1/ cos t cos t dt . = dt = (sin t/ cos t)2 sin2 t Now let w = sin t, so that dw = cos t dt: Z dw 1 = = −w−1 + C = − + C. w2 sin t Finally, we need to express this in terms of x again. Since tan t = x, we can draw the following triangle: x t 1 √ 2 The length of the √ hypoteneuse of this triangle must be 1 + x , and this tells us that sin t = x/ 1 + x2 . So we can finish as follows: √ Z 1 + x2 1 1 √ +C = − +C. dx = − sin t x x2 1 + x2 3. The aquarium contains 4 cubic meters of water, so we only need to pump out the top quarter of the water. Consider a layer of water of thickness ∆y at height y meters above the bottom. The mass of this layer is 3 (density) × (volume) = 1000 kg./m × 4 × 1 × ∆y m3 = 4000∆y kg. The force needed to raise this layer is 9.8 × 4000∆y newtons. This layer must be raised a distance of 1 − y meters. Since we are only moving water from the top quarter of the tank, y ranges from 3/4 to 1. Therefore, the work (in joules) required for all of the layers is about X 9.8 × 4000 × ∆y × (1 − yi ) i (where yi is the height of the ith layer). This corresponds to the integral Z 1 9.8 × 4000(1 − y) dy . 3/4 4. (This is similar to Example 4 on page 426 of the text; see the sketches there.) The two curves intersect at √ (0, 0) and (1, 1). Between these two points (i.e. for 0 ≤ x ≤ 1), the curve y = x lies above the curve y = x4 (and both lie above the x-axis). Consider the solid obtained by rotating this region about the x-axis. For each number x between 0 and 1, let A(x) be the cross-section area of the slice through the solid by the plane that is perpendicular to the x-axis and that intersects the x-axis at the √ given value of x. The shape of the cross-section is a “washer” with outer radius x and inner radius x4 (i.e., the radius of the hole in the washer is x4 ). Therefore √ A(x) = π( x)2 − π(x4 )2 = π(x − x8 ) and the volume of the solid is 2 Z 1 Z 1 x x9 − A(x) dx = π(x − x8 ) dx = π 2 9 0 0 1 = π 0 1 1 − 2 9 = 7 π. 18 5. (a) We have ∆x = 2 4−2 = , x0 = 2, 3 3 x1 = 8 10 , x2 = , x3 = 4 3 3 so the Trapezoidal Rule estimate is T3 = = = = ∆x (f (x0 ) + 2f (x1 ) + 2f (x2 ) + f (x3 )) 2 2/3 (ln(2) + 2 ln(8/3) + 2 ln(10/3) + ln(4)) 2 1 (.6931 + 2(.9808) + 2(1.2040) + 1.3863) 3 2.1493. (b) The absolute value of the error is less than or equal to |4 − 2|3 K2 , 12n2 where K2 is a number that satisfies d2 ln x ≤ K2 dx2 for every x in [2, 4]. Since the second derivative of ln x is −x−2 , we need to make sure that x−2 ≤ K2 for every x in [2, 4]. Since the largest that x−2 gets (for x in [2, 4]) is 2−2 , we can take K2 = 2−2 = 0.25. So to guarantee that Tn is accurate to 4 decimal places, we need to choose n large enough so that |4 − 2|3 (0.25) ≤ 10−4 12n2 i.e. i.e. 2 × 104 ≤ n2 12 40.82 ≤ n Therefore, 41 subintervals suffice. (c) The function ln x is concave down (note that its second derivative is negative). A sketch of a positive concave-down function shows that the trapezoids would all lie below the graph of the function. Therefore the sum of the areas of the trapezoids is smaller than the area under the curve. For this reason, we expect the estimate from (a) to be a bit smaller than the true value of the integral.