QUIZ #13: CALCULUS 1A (Stankova) Wednesday, April 28, 2004 Section 10:00–11:00 (Voight)

Problem 1. Let g(x) = ∫_{−2}^{x} f(t) dt, where f is the function shown.

(a) Evaluate g(−2).

(b) Is g(4) > 0? Explain.

(c) Estimate g(0).

(d) Where does g have a maximum value in the interval [−2, 4]?

(e) Draw a (very) rough graph of g.

Solution. For (a), we have g(−2) = 0 (an integral with endpoints equal is zero). For (b), yes, g(4) > 0 because the total area above the axis is greater than that below. For (c), we estimate using right endpoints and the intervals [−2, −1] and [−1, 0] (with Δx = 1), which has f(−1) = 3 and f(0) = 1 so

g(0) ≈ ∑_{i=1}^{2}f(x_{i}) Δx = (3 + 1)(1) = 4.

For (d), g has a maximum value at x = 4, since this is where the total area will be maximal. The plot of g looks like:

QUIZ #13: CALCULUS 1A (Stankova) Wednesday, April 28, 2004 Section 11:00–12:00 (Voight)

Problem 1. Use the Fundamental Theorem of Calculus (Part 1) to find the derivative of the function

y = ∫_{1}^{x^{2}}(\sqrt{t} + ln t) dt.

Solution. If

y = ∫_{1}^{h(x)} f(t) dt

then

y′ = f(h(x))h′(x).

So since f(t) = \sqrt{t} + ln t and h(x) = x^{2}, we have

y′ = (\sqrt{x^{2}} + ln(x^{2}))(2x) = 2x|x| + 4x ln x.

Problem 2. Use the Fundamental Theorem of Calculus (Part 2) to evaluate the integral.

∫_{1}^{2} \frac{6 + \sqrt{u}}{u^{2}} du.

Solution. The antiderivative of

f(u) = \frac{6 + \sqrt{u}}{u^{2}} = 6u^{−2} + u^{−3/2}

is

F (u) = −6u^{−1} − 2u^{−1/2} = −\frac{6}{u} − 2\frac{1}{\sqrt{u}}.

By the FTC we have

∫_{1}^{2} \frac{6 + \sqrt{u}}{u^{2}} du = F (2) − F (1) = −3 − \frac{2}{\sqrt{2}} − (−6 − 2) = 5 − \sqrt{2}.