Mathematics 134 – Intensive Calculus for Science 2 Exam I Solutions – February 17, 2006

I. The given graph shows y = f(x).

A) (10) Use the graph to determine ∫_{0}^{4} f(x) dx.

Solution: Using the interpretation of the integral as signed area, we see there is a triangle below the x-axis. The region above the x-axis can be split into a small triangle, a trapezoid (2 ≤ x ≤ 3) and another larger triangle:

∫_{0}^{4} f(x) dx = −\frac{1}{2} · \frac{3}{2} · 1 + \frac{1}{2} · \frac{1}{2} · 1 + \frac{1 + 2}{2} · 1 + \frac{1}{2} · 1 · 2 = 2.

B) (5) Let F (x) be the antiderivative of f(x) with F (0) = −2. What is F (4)?

Solution: By the Fundamental Theorem of Calculus

2 = ∫_{0}^{4} f(x) dx = F (4) − F (0) = F (4) + 2 ⇒ F (4) = 0

C) (15) Give a qualitative sketch of y = F (x) from part B. Your graph should show where F is increasing/decreasing, and where F is concave up/concave down.

Solution: Proceeding as in part B we can get the following values: F (0) = −2, F (1) = −5/2, F (2) = −5/2, F (3) = −1 and F (4) = 0. The graph of F is decreasing on 0 < x < 3/2, and increasing the rest of the way to x = 4. It is concave down for 0 < x < 1 and again for 3 < x < 4. It is concave up for 1 < x < 3. The graph looks like this:

D) (5) The function G(x) is G(x) = ∫_{0}^{x} f(t) dt using the function f in the graph above. What is G′(2)?

Solution: By the second part of the Fundamental Theorem, G′(2) = f(2) = 1, from the given graph.

II. On the Moon, the acceleration due to gravity is about -1.6 m/sec^{2} (versus about -9.8 m/sec^{2} on Earth).

A) (10) A rock is tossed upward with initial velocity v(0) = 1 m/sec. Solve the equation \frac{dv}{dt} = −1.6 to find the velocity of the ball as a function of time.

Solution: v(t) = ∫ −1.6 dt = −1.6t + C. From the initial condition v(0) = 1, C = 1, so v(t) = −1.6t + 1.

B) (10) Use your answer v(t) from part A. If the rock’s initial height is h(0) = 1 m, solve the equation \frac{dh}{dt} = v(t) to find the height of the rock as a function of time.

Solution: The height is the integral of the velocity:

h(t) = ∫ v(t) dt = ∫ −1.6t + 1 dt = −0.8t^{2} + t + C

From the initial height h(0) = 1, we get C = 1, so h(t) = −0.8t^{2} + t + 1. C) (10) How long does the rock take to fall back to ground level on the Moon? Would

the corresponding time be longer or shorter on the Earth?

Solution: We solve the equation h(t) = 0 by the quadratic formula:

t = \frac{1 ± \sqrt{1 − 4(−.8)(1)}}{2(−.8)} ≐ −.66, 1.91

We take the positive value 1.91 sec. This is longer than the corresponding time on Earth. (Intuitively, the Moon’s gravity is “weaker” and pulls the rock down more slowly.)

III. Compute each of the integrals below using basic rules and/or substitution. You must show all work for full credit.

A) (10)

∫ 3x^{7} − \frac{4}{x^{3}} + 5\sqrt{x} + \frac{7}{x} dx

Solution: By the power rule (including the special case for ∫ x^{−1} dx)

= \frac{3}{8}x^{8} + 2x^{−2} + \frac{10}{3}x^{3/2} + 7 ln |x| + C

B) (15)

∫_{0}^{1} sin(πx) + cos(πx) dx

Solution: We apply the substitution u = πx, so du = πdx. The integral becomes

\frac{1}{π} ∫_{u=0}^{u=π} sin(u) + cos(u) du = \frac{1}{π} (− cos(u) + sin(u)|_{0}^{π}

= \frac{1}{π} (− cos(π) + sin(π) − (− cos(0) + sin(0)))

= \frac{2}{π}

C) (10)

∫ xe^{5x^{2}+7} dx

Solution: Use u = 5x^{2} + 7, so du = 10xdx. The integral is

\frac{1}{10} ∫ e^{u} du = \frac{1}{10}e^{u} + C = \frac{1}{10}e^{5x^{2}+7} + C