College of the Holy Cross, Spring 2014 Math 136, section 1, Midterm Exam 1 Friday, February 21

I. Let f(x) = x^{3} + 1 on the interval [a, b] = [1, 3].

A. (10) Evaluate the Riemann sum for f on this interval using n = 4 and x_{i}^{∗} = right endpoints.

Solution: With n = 4, we have Δx = \frac{3−1}{3} = \frac{1}{2}. The end points of the intervals are x_{0} = 1, x_{1} = \frac{3}{2}, x_{2} = 2, x_{3} = \frac{5}{2}, and x_{4} = 3. The right-hand Riemann sum equals:

f(3/2)Δx + f(2)Δx + f(5/2)Δx + f(3)Δx,

which equals

(1/2)[((3/2)^{3} + 1) + (2^{3} + 1) + ((5/2)^{3} + 1) + (3^{3} + 1)] = 29

B. (10) Now repeat part A, but using the left endpoints.

Solution: Similar to part A, but using the left endpoints:

f(1)Δx + f(3/2)Δx + f(2)Δx + f(5/2)Δx,

which equals

(1/2)[(1^{3} + 1) + ((3/2)^{3} + 1) + (2^{3} + 1) + ((5/2)^{3} + 1)] = 16

C. (5) One of your answers in parts A and B is definitely an overestimate for the value of ∫_{1}^{3} x^{3} + 1 dx. Which is it?

Solution: Since f(x) = x^{3} + 1 is increasing on its whole domain, the right-hand sum (part A) gives an overestimate of the integral. (The exact value is

∫_{1}^{3} x^{3} + 1 dx = \frac{x^{4}}{4} + x|_{1}^{3} = \frac{81}{4} + 3 − \frac{1}{4} − 1 = 22.)

II. Compute the derivatives of each of the following functions defined by integrals.

A. (5) f(x) = ∫_{1}^{x} e^{t^{2}} dt

Solution: By the FTC, part 1, f′(x) = e^{x^{2}}

B. (5) g(x) = ∫_{x^{2}}^{4} \frac{cos(t)}{t^{2}} dt

Solution: We must reverse the order of the limits, then use the FTC part 1 and the Chain Rule on this one:

g(x) = − ∫_{4}^{x^{2}} \frac{cos(t)}{t^{2}} dt ⇒ g′(x) = \frac{− cos(x^{2})}{x^{4}} · 2x = \frac{−2 cos(x^{2})}{x^{3}}.

III. The following graph (made up of straight line segments) shows y = f(t) for 0 ≤ t ≤ 4.

Given: f(1) = 2, f(2) = −2, f(3) = −1, and f(4) = 0. The function F is defined by F (x) = ∫_{0}^{x} f(t) dt.

A. (5) Determine the values F (x) for x = 0, 1, 2, 3, 4 and enter them in the following table.

Solution: By finding areas of trapezoids and triangles,

\begin{matrix}x & 0 & 1 & 2 & 3 & 4 \\ F (x) & 0 & 3/2 & 3/2 & 0 & -1/2 \end{matrix}

B. (5) Does F (x) have any critical points? If so, say where. If not say why not.

Solution: F′(x) = f(x) and this equals 0 when x = 3/2 and (in a sense – see below) again when x = 4. Of these x = 3/2 is definitely a critical point. Whether or not you call x = 4 a critical point is sort of a matter of taste (i.e. of the precise way critical points are defined.) Note that F (x) is not defined for x > 4, so to define F′(4), one would have to consider just a one-sided limit of the difference quotient. That does exist and equals zero here. So it is OK to say F′(4) = 0 in that sense.

C. (5) Over which interval(s) is F (x) concave down?

Solution: This is true on intervals where F′(x) = f(x) is decreasing, so (1, 2)

IV.

A. (5) Integrate with a suitable u-substitution: ∫_{0}^{1}(4x^{3} + 1)^{3/5}x^{2} dx.

Solution: Let u = 4x^{3} + 1. Then du = 12x^{2} dx, so \frac{1}{12}du = x^{2} dx. When x = 0, u = 1 and when x = 1, u = 5. The definite integral goes over to

\frac{1}{12} ∫_{1}^{5} u^{3/5} du = \frac{1}{12} \frac{5}{8} u^{8/5}|_{1}^{5} = \frac{5}{96}[5^{8/5} − 1] ≐ .6319

B. (10) Integrate with a suitable u-substitution: ∫ \frac{x sin(3x^{2})}{cos(3x^{2}) + 1} dx.

Solution: Let u = cos(3x^{2}) + 1. Then du = −6x sin(3x^{2}) dx. So x sin(3x^{2}) dx = \frac{−1}{6} du and the integral goes over to

\frac{−1}{6} ∫ \frac{du}{u} = \frac{−1}{6} ln |u| + C = \frac{−1}{6} ln | cos(3x^{2}) + 1| + C.

C. (15) Integrate by parts: ∫ x^{2} sin(5x) dx

Solution: We integrate by parts twice, letting u = power of x each time:

∫ x^{2} sin(5x) dx = \frac{−x^{2} cos(5x)}{5} + \frac{2}{5} ∫ x cos(5x) dx
= \frac{−x^{2} cos(5x)}{5} + \frac{2}{5} (\frac{x sin(5x)}{5} − \frac{1}{5} ∫ sin(5x) dx)
= \frac{−x^{2} cos(5x)}{5} + \frac{2x sin(5x)}{25} + \frac{2 cos(5x)}{125} + C.

D. (10) Integrate with any applicable method we have discussed: ∫_{0}^{1} \frac{x}{\sqrt{x^{2} + 1}}dx

Solution: The x dx in the numerator is, up to a constant, the derivative of u = x^{2} + 1 in the radical:

= \frac{1}{2} ∫_{u=1}^{u=2} u^{−1/2} du = u^{1/2}|_{1}^{2} = \sqrt{2} − 1.

E. (10) Integrate with any applicable method we have discussed: ∫ e^{x} cos(4x) dx

Solution: Use parts twice (letting dv = e^{x} dx each time), then solve for the integral:

∫ e^{x} cos(4x) dx = e^{x} cos(4x) + 4 ∫ e^{x} sin(4x) dx
= e^{x} cos(4x) + 4 (e^{x} sin(4x) − 4 ∫ e^{x} cos(4x) dx)
so ∫ e^{x} cos(4x) dx = \frac{e^{x}}{17}(cos(4x) + 4 sin(4x)) + C.