Math 116- 101 or 102 (Circle One)  Name:  \underline{SOLUTIONS}

Quiz 2: §7.1,7.2,7.5 05/15/2012

Show all work and include units where appropriate. You have 30 minutes to complete this quiz. (25 pts)

1. Let g be a function differentiable on 0 ≤ x ≤ 4. Suppose ∫_{0}^{4} g(x)dx = 10. A table of values of g and g′ is given below. Compute the following integrals exactly. Show all work.

x  0 1 2 3 4
g(x) -1 2 6 2 0
g’(x) 2 3 -1 -3 1

(a) ∫_{0}^{4} xg′(x)dx. (3 pts)

Using integration by parts with u = x (and thus du = dx) and dv = g′(x)dx (and thus v = g(x)), we have ∫ xg′(x)dx = xg(x) − ∫ g(x)dx. Therefore ∫_{0}^{4} xg′(x)dx = xg(x)|_{0}^{4} − ∫_{0}^{4} g(x)dx = [4g(4) − 0g(0)] − ∫_{0}^{4} g(x)dx = [4 ∗ 0 − 0 ∗ (−1)] − 10 = −10.

(b) Compute ∫_{1}^{2} xg′(x^{2})dx. (3 pts)

Using a substitution with w = x^{2} (and thus dw = 2xdx), we obtain ∫_{1}^{2} xg′(x^{2})dx = ∫_{1^{2}}^{2^{2}} \frac{1}{2}g′(w)dw = \frac{1}{2}g(w)|_{1}^{4} = \frac{1}{2}[g(4) − g(1)] = \frac{1}{2}[0 − 2] = −1

(c) Compute ∫_{3}^{4} g′(x)g′(g(x))dx (3 pts)

Using a substitution with w = g(x) (and thus dw = g′(x)dx), we obtain ∫_{3}^{4} g′(x)g′(g(x))dx = ∫_{g(3)}^{g(4)} g′(w)dw = g(w)|_{2}^{0} = g(0) − g(2) = −1 − 6 = −7

(d) ∫_{0}^{2} x^{3}g′(x^{2})dx (3 pts) [HINT: It may help to use your answer from part (a).]

Using a substitution with w = x^{2} (and thus dw = 2xdx), we obtain ∫_{0}^{2} x^{3}g′(x^{2})dx = ∫_{0^{2}}^{2^{2}} \frac{1}{2}wg′(w)dw = \frac{1}{2} ∫_{0}^{4} wg′(w) = \frac{1}{2}(−1) = \frac{−1}{2}, where the penultimate equality comes from part (a).

2. Will and James are kayaking in the Gulf of Mexico. Their velocity, v(t), (in \frac{miles}{hour}) is a function of how long, t, (in hours) they have been kayaking. v(t) is given by the function v(t) = \frac{−1}{6}t^{3} − \frac{1}{2}t^{2} + 4t.

(a) Is v(t) increasing or decreasing on the interval 0 ≤ t ≤ 2? Justify your answer. (2 pts)

We note v′(t) = \frac{−1}{2}t^{2} − t + 4 ≥ 0 for 0 ≤ t ≤ 2, thus v(t) is increasing on 0 ≤ t ≤ 2.

(b) Is v(t) concave up or concave down on the interval 0 ≤ t ≤ 2? Justify your answer. (2 pts)

We note v′′(t) = −t − 1 < 0 for 0 ≤ t ≤ 2, thus v(t) is concave down on 0 ≤ t ≤ 2

(c) Using a left hand Riemann sum with n = 4 subintervals, estimate the distance Will and James kayaked during those two hours. Include units. (2 pts)

Since a = 0, b = 2, n = 4, we have Δt = \frac{2−0}{4} = .5. Thus ∫_{0}^{2} v(t)dt ≈ LEF T (4) = Δt[v(0) + v(.5) + v(1) + v(1.5)] = .5[0 + 1.85 + 3.33 + 4.31] ≈ 4.75 miles.

(d) Using the midpoint rule with n = 4 subintervals, estimate the distance Will and James kayaked. Include units. (3 pts)

Since a = 0, b = 2, n = 4, we have Δt = \frac{2−0}{4} = .5. Thus ∫_{0}^{2} v(t)dt ≈ MID(4) = Δt[v(.25) + v(.75) + v(1.25) + v(1.75)] ≈ .5[.96 + 2.64 + 3.89 + 4.58] = 6.04 miles.

(e) Without computing the exact distance Will and James kayaked, are your answers from parts (c) and (d) overestimates or underestimates? Justify. (4 points)

Because v(t) is increasing for 0 ≤ t ≤ 2, we have LEF T (4) < ∫_{0}^{2} v(t)dt. In other words, the left hand Riemann sum is an UNDERSETIMATE. Because v(t) is concave down for 0 ≤ t ≤ 2, we have MID(4) > ∫_{0}^{2} v(t)dt. In other words, the midpoint rule gives an OVERESTIMATE.