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Math 113, Calculus II Test 1: Review Solutions (g) ∫ ln x 3 dx x 3 Solution: While substitution may look tempting, it isn’t what we want for this one. Instead, after simplifying ln(x 3 ) = 3 ln x, we see that ∫ ln x 3 x 3 dx = ∫ 3 ln x x 3 dx = 3 ∫ ln x x 3 dx is just 3 times the integral we computed in part (d) using PARTS (u = ln x and dv = x −3 dx). (h) ∫ (ln x) 3 dx Solution: For this one, we have to use PARTS with u = (ln x) 3 (so du = 3(ln x) 2 1 x dx) and dv = dx (so v = x). Then ∫ ∫ (ln x) 3 dx = x(ln x) 3 − x · 3(ln x) 2 1 ∫ x dx = x(ln x)3 − 3 (ln x) 2 dx and we see that we’ve decreased the degree of ln x in the integral by one in the process so the result of using parts is simpler than what we started with. Doing parts again on ∫ (ln x) 2 dx we would decrease the degree by one more, and another application of parts gets rid of the ln x completely. 9. Let f be a continuous function with antiderivative F on the interval [a, b]. Let c be any point in the interval. State whether the following are true or false. If false, then give an example to show why it is false and correct the statement if possible If true, explain why. (a) ∫ b a f(x)dx = ∫ c a f(x)dx = ∫ b c f(x)dx Solution: FALSE! If c = a, then ∫ c a f(x)dx = 0 but certainly ∫ b a f(x)dx isn’t always zero. The “mistake” was made by writing an equality sign rather than addition on the right: ∫ b a f(x)dx = ∫ c a f(x)dx + ∫ b c f(x)dx. (b) ∫ b a F (x)dx = f(b) − f(a) Solution: FALSE! This is similar to the Evaluation Theorem, which says ∫ b a f(x)dx = F (b) − F (a), but it makes the “mistake” of exchanging the roles of F and f. (c) ∫ b a f(x)dx ≥ 0 Solution: FALSE! If f(x) < 0 for all x in [a, b], say f(x) = −1, then ∫ b a f(x)dx will be < 0 (in the example of f(x) = −1, we have ∫ b a −1dx = (−1)(b − a), which is negative if b > a). (d) ∫ b a cf(x)dx = c[F (b) − F (a)] Solution: TRUE! First, ∫ b then ∫ b a a cf(x)dx = c ∫ b a f(x)dx by the Constant Multiple Rule and f(x)dx = F (b) − F (a) by the Evaluation Theorem. (e) ∫ b a f(x)dx = f(m)(b − a) for some m in [a, b]. Solution: TRUE! This is the Mean Value Theorem for Integrals. 10. Let A(t) be the area under the curve y = sin(x 2 ) from 0 to t and B(t) be the area of the triangle with vertices at O = (0, 0), P = (t, sin(t 2 )), and (t, 0). Find lim t→0 + A(t)/B(t).
Math 113, Calculus II Test 1: Review Solutions Solution: The value of B(t) is the area of the triangle, which is one-half its base (t) times its height (sin(t 2 )), so B(t) = 1 2 t sin t2 while A(t) = ∫ t 0 sin x2 dx. To compute the limit, we’re going to use l’Hôpital’s Rule several times and the Fundamental Theorem of Calclus in our first application of it: A(t) lim t→0 + B(t) = lim 1 2 ∫ t t→0 + t→0 + ∫ t = 2 lim 0 sin x2 dx t sin t2 0 sin x2 dx t sin t 2 = 2 lim t→0 + sin t 2 sin t 2 + t cos(t 2 )[2t] cos(t 2 )[2t] = 2 lim t→0 + cos(t 2 )[2t] + [4t] cos(t 2 ) + [2t 2 ][− sin(t 2 )[2t]] = 2 lim t→0 + [2] cos(t 2 ) + [2t][− sin(t 2 )[2t]] [6] cos(t 2 ) + t(· · · ) = 2 [2] + 0 [6] + 0 = 2 3 . 11. Consider the function g(x) = ∫ 3 x f(t)dt where the function f(t) is given by the graph on your review sheet. (a) On what intervals is g increasing? Decreasing? Solution: The function g is increasing when its derivative g ′ (x) is positive and is decreasing when g ′ (x) is negative. So we need to find g ′ (x), but g(x) = ∫ 3 x f(t)dt = − ∫ x 3 f(t)dt, so the Fundamental Theorem of Calculus (Part I) implies that g ′ (x) = d ∫ x dx [− f(t)dt] = − d ∫ x dx [ f(t)dt] = −f(x). 3 Hence, g is increasing when g ′ (x) = −f(x) > 0, i.e., when f(x) < 0, so g is increasing on the interval (2, 4). Likewise, g is decreasing when g ′ (x) = −f(x) < 0, i.e., when f(x) > 0, so g is decreasing on the intervals (0, 2) and (4, 6). (b) At what values of x does g have a local max? Local min? Solution: The function g has a local max where it changes from increasing to decreasing (by the First Derivative Test), which occurs at x = 4. The function g has a local min where it changes from decreasing to increasing, which happens at x = 2. (c) On what intervals is g concave up? Concave down? Solution: The function g is concave up when its second derivative g ′′ (x) = d dx [g′ (x)] = d dx [−f(x)] = −f ′ (x) is positive, so g is concave up when f ′ (x) < 0. That is, g is concave up when the graph of y = f(x) is decreasing, which is the case on the interval (1, 3). The function g is concave down where the graph of f is increasing (so that f ′ (x) > 0), which occurs on the intervals (0, 1) and (3, 5). 12. Show that the area enclosed by the graph of the parabola 3 y = f(x) = 2 a 2 x − 1 a 3 x2 (a > 0) and the x-axis has an area independent of a. How large is the area?
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