Solutions to Review Questions, 5.3-5.5 - Answer Key

Solutions to Review Questions, 5.3-5.5 1. True or False, and give a short reason: Z b (a) If f and g are continuous on [a, b], then Z b f (x) + g(x) dx = Z f (x) dx + a a True. This is one of the properties of definite integrals. Z b Z (b) If f and g are continuous on [a, b], then f (x)g(x) dx = a b g(x) dx a Z f (x) dx · a b b g(x) dx a Oooh- this is bad on so many levels! We can use just about any functions f and g and this will not work. For example, Z 1 x · x dx = x3 + C 3
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R R 1 2 but x dx · x dx = 2 x + C1 2 x + C2 Another explanation would be that since d (f (x) · g(x)) 6= f 0 (x) · g 0 (x) dx the reverse does not work, either. Z (c) If f is continuous on [a, b], then b Z xf (x) dx = x a b f (x) dx a False. The result of doing this R R would give you R a function of x, but Rthe definite integral is a number. It is also false in general, xf (x) dx 6= x f (x) dx; for example x dx = 12 x2 + C, but x dx = x2 + Cx Z 4 (d) If f 0 is continuous on [−1, 4], then f 0 (w) dw = f (4) − f (−1) −1 This is true, since the antiderivative of f 0 is f (that is, the derivative of f is f 0 ). So, this is the Fundamental Theorem of Calculus, part II. Z 1 1 3 (e) dx = − 4 x 8 −2 False. We cannot use the FTC on this integral since x14 has a vertical asymptote at x = 0, which is inside the interval [−2, 1]. (f) All continuous functions have derivatives. False. For example, y = |x| does not have a derivative at x = 0, although it is continuous there. (g) All continuous functions have antiderivatives. True. This is the Fundamental Theorem of Calculus, Part I written out in words. Z 7 (h) If v(t) is velocity at time t, then the distance traveled between times 3 and 7 is given by v(t) dt 3 Z 7 False. To get the total distance traveled, we would evaluate |v(t)| dt. If we want the relative Z 7 3 distance, we would ask for displacement, which is given as v(t) dt 3 (i) Even though the function: x2 3+x  f (x) = if x < 1 if x > 1 2 Z is not continuous at x = 1, we can compute f (x) dx. 0 This is true. Because there is not a vertical asymptote at x = 1, we can compute the integral by breaking it up: Z 2 Z 1 Z 2 f (x) dx = x2 dx + 3 + x dx 0 0 1 1 2. Compare the notation: Z x d f (t) dt This value is f (x) (a) dx a Z b d (b) f (t) dt This value is zero (it’s the derivative of a constant). dx a Z b d (c) f (x) dx This value is f (b) − f (a) dx a Z b (d) f (x) dt This is a little tricky- since the integral is in terms of t, but f is an expression in x, it a will be constant with respect to t. The answer: f (x)(b − a). 3. Evaluate, where possible. If not, state why: Z 4 Z 4 2 4 2 146 x −x+1 2 √ dx = x3/2 − x1/2 + x−1/2 dx = x5/2 − x3/2 + 2x1/2 = (a) 5 3 15 x 1 1 1 Z 2 x (b) dx The function has a vertical asymptote at x = 1; FTC does not apply. 2 − 1)2 (x 0 Z 3x d (c) t dt = 3x · 3x ln(3) = 32x ln(3) dx 3 Z p (d) (1 − x) 2x − x2 dx Use u = 2x − x2 , du = 2 − 2x dx, or 12 du = 1 − x dx. This gives: Z Z (e) Z (f) 0 Z Z p 1 1 2 1 (1 − x) 2x − x2 dx = u1/2 du = · u3/2 + C = · (2x − x2 )3/2 + C 2 2 3 3 cos(ln(x)) dx Use u = ln(x), so du = x1 dx, which gives: x Z Z cos(ln(x)) dx = cos(u) du = sin(u) + C = sin(ln(x)) + C x 1 d dx  ex x+1  dx = ex x+1 1 = 0 1 e−1 2 2π | sin(x)| dx The sine function is positive for 0 < x < π and negative for π < x < 2π, so we (g) 0 rewrite: Z π Z 2π sin(x) dx − 0 Z (h) (i) π π 2π = − cos(x)|0 + cos(x)|π = 2 + 2 = 4 x dx You might have tried u = 1 − x4 , but that doesn’t get us very far. Note that this 1 − x4 1 expression is close to √1−x , whose antiderivative is sin−1 (x). This suggests that we try: u = x2 , 2 du = 2x dx or 21 du = x dx. Putting this together, Z Z x 1 1 1 1 √ √ dx = du = sin−1 (u) + C = sin−1 (x2 ) + C 2 2 2 1 − x4 1 − u2 d dx √ Z 3x+1 sin(t4 ) dt Here we use the general formula: 2x rewrite: 4 Z 2 1 Z h(x) f (t) dt = f (h(x)) · h0 (x). First, a Z 3x+1 d d 4 sin(t ) dt = sin(t ) dt + sin(t4 ) dt = dx 2x dx 1 2x Z 2x Z 3x+1 d d − sin(t4 ) dt + sin(t4 ) dt = − sin((2x)4 ) · 2 + sin((3x + 1)4 ) · 3 dx 1 dx 1 d dx Z 3x+1 d dx x2 dx Here, we’ll try u = 1 − x, so that du = −dx. But we still have x2 in the numerator. 1−x Use u = 1 − x so that x = 1 − u and x2 = (1 − u)2 = 1 − 2u + u2 , and now do the full substitution: Z Z Z 1 − 2u + u2 x2 √ √ dx = − du = − (u−1/2 − 2u1/2 + u3/2 ) du = u 1−x   √ 2 2 4 2 − 2u1/2 − 2 · u3/2 + u5/2 + C = −2 1 − x + (1 − x)3/2 − (1 − x)5/2 + C 3 5 3 5 Z (j) √ Note: Since C is arbitrary, you could have either C or −C in your answer Z 4 Z 2 4. If f is continuous and f (x) dx = 10, find f (2x) dx With that 2x, we should be tempted to use u = 2x, du = 2 dx or says: 0 1 2 du 0 = dx. Now, for the bounds, if x = 0, then u = 0 and if x = 2, u = 4. This 2 Z 1 2 f (2x) dx = 0 Z Z 4 f (u) du = 0 1 · 10 = 5 2 x 1 dt, find where g is concave up. To find where g is concave up, we must compute 1 + t + t2 0 its second derivative and find where it is positive: 5. If g(x) = g 0 (x) = 1 = (1 + x + x2 )−1 1 + x + x2 g 00 (x) = −(1 + 2x) (1 + x + x2 )2 Since the denominator is always positive, we need look only for where the numerator is positive: −1 − 2x > 0 ⇒ −2x > 1 ⇒ x < If x < 21 , then g 00 (x) > 0, so g is concave up. Z 6 Z 4 Z 6. If f (x) dx = 10 and f (x) dx = 7, find 0 0 1 2 6 f (x) dx. By a property of integrals, 4 Z 6 Z 6 4 Z f (x) dx − f (x) dx = 4 0 f (x) dx = 10 − 7 = 3 0 Challenge Problems! If you breezed through the previous questions, and are looking for something more challenging, try these! 1. d2 dx2 Z First, let F (t) = sin(t) Z 0 x Z sin(t) ! p 1 + u4 du dt 1 p 1 + u4 du, so that the first derivative is: 1 d dx Z x Z F (t) dt = F (x) = 0 sin(x) p 1 + u4 du 1 Now the next derivative is our standard problem: d dx Z sin(x) p p 1 + u4 du = 1 + (sin(x))4 · cos(x) 1 3 Z 2. If f is a differentiable function so that: x 2 f (t) dt = (f (x)) for all x, find f . 0 We can remove the integral by differentiating both sides: f (x) = 2f (x) · f 0 (x) ⇒ f (x) − 2f (x) · f 0 (x) = 0 ⇒ f (x) (1 − 2f 0 (x)) = 0 0 This says that either f (x) R x = 0 or 1 − 2f (x) = 0. For the first case, we see that f (x) = 0 will solve our original function, since 0 0 dx = 0 for all x. In the second case, f 0 (x) = 12 , so f (x) = 12 x + C. To get the value of C, notice in the original equation that if x = 0, then: Z 0 f (x) dx = (f (0))2 ⇒ f (0) = 0 0 Thus, C = 0. So, we have two possibilities: f (x) = 0 or f (x) = 21 x. 3. Find 1 lim h→0 h 2+h Z p 1 + t3 dt 2 x Z Do we recognize this as a derivative? If g(x) = f (t) dt, then 0 1 1 g 0 (2) = lim (g(2 + h) − g(2)) = lim h→0 h h→0 h x Z In particular, if g(x) = Z 2+h Z f (t) dt − 0 2 ! f (t) dt 0 1 = lim h→0 h Z 2+h f (t) dt 2 p √ 1 + t3 dt, then g 0 (x) = 1 + t3 , so 0 1 h→0 h Z lim 2+h p p √ 1 + t3 dt = g 0 (2) = 1 + 23 = 9 = 3 2 Note: the use of 0 in the definition of g was arbitrary- we could’ve used any constant greater than −1. 4