Mathematics 134 – Intensive Calculus for Science 2 Solutions – Exam 2 March 25, 2002

I. A) Integrate by parts with u = x, dv = cos(4x) dx. Then du = dx and v = ∫ cos(4x) dx (let w = 4x; then form is v = \frac{1}{4} ∫ cos(w) dw = \frac{1}{4} sin(4x). Then by the parts formula:

∫ x cos(4x) dx = \frac{x}{4} sin(4x) − ∫ \frac{1}{4} sin(4x) dx

= \frac{x}{4} sin(4x) + \frac{1}{16} cos(4x) + C

B) To use parts here, you want to think of splitting the x^{3} into x^{2} · x, and then put the x into dv to make the dw to evaluate ∫ x(x^{2} + 12)^{\frac{1}{2}} dx by the substitution w = x^{2} + 12. So

u = x^{2}  dv = ∫ x(x^{2} + 12)^{\frac{1}{2}} dx
du = 2x dx v = \frac{1}{2} ∫ w^{\frac{1}{2}} dw
=  \frac{1}{2} · \frac{2}{3}w^{\frac{3}{2}}
=  \frac{1}{3}(x^{2} + 12)^{\frac{3}{2}}

Hence by the parts formula,

∫ x^{3}(x^{2} + 12)^{\frac{1}{2}} dx = \frac{x^{2}}{3} (x^{2} + 12)^{\frac{3}{2}} − \frac{2}{3} ∫ x(x^{2} + 12)^{\frac{3}{2}} dx

= \frac{x^{2}}{3} (x^{2} + 12)^{\frac{3}{2}} − \frac{2}{3} · \frac{1}{2} · \frac{2}{5}(x^{2} + 12)^{\frac{5}{2}} + C

= \frac{x^{2}}{3} (x^{2} + 12)^{\frac{3}{2}} − \frac{2}{15}(x^{2} + 12)^{\frac{5}{2}} + C

(doing the same substitution used to get v).

II. A) The quadratic equation x^{2} + 8x + 17 = 0 has no real roots. Hence we want to complete the square and use table entry # 24:

∫_{0}^{1} \frac{dx}{x^{2} + 8x + 17} = ∫_{0}^{1} \frac{dx}{(x + 4)^{2} + 1}

= arctan(x + 4)|_{0}^{1}

= arctan(5) − arctan(4)

B) This is a polynomial times an exponential, so table entry # 14 is the one to use, with p(x) = x^{3} + x and a = 2. Using the second line in the table entry, which shows the general result when the first line is applied repeatedly, we have:

\begin{aligned}∫ (x^{3} + x)e^{2x} dx & = \frac{1}{2}(x^{3} + x)e^{2x} − \frac{1}{4}(3x^{2} + 1)e^{2x} + \frac{1}{8}(6x)e^{2x} − \frac{1}{16}(6)e^{2x} + C \\ & = \frac{1}{2}(x^{3} + x)e^{2x} − \frac{1}{4}(3x^{2} + 1)e^{2x} + \frac{3}{4}xe^{2x} − \frac{3}{8}e^{2x} + C \end{aligned}

III. B) Rotating about the x-axis, the slices are circular disks, so

volume of slice ≐ πr^{2}Δx

= π(x + 2)^{2}Δx

total volume ≐ ∑ π(x + 2)^{2}Δx

= ∫_{0}^{1} π(x + 2)^{2} dx

= π ∫_{0}^{1} x^{4} + 4x^{2} + 4 dx

= π (\frac{x^{5}}{5} + \frac{4x^{3}}{3} + 4x) |_{0}^{1}

= \frac{83π}{15}

C) Rotating about y = 5, the slices are circular “washers”, since there is now a gap between the region we are rotating and the axis of rotation. The outer radius is r_{out} = 5 (constant) because it is always 5 units from the line y = 5 to the bottom edge of the region along y = 0. The inner radius is r_{in} = 5 − (x^{2} + 2) = 3 − x^{2} (vertical separation between the line y = 5 and the graph). Hence the volume is

∫_{0}^{1} π5^{2} − π(3 − x^{2})^{2} dx.

IV. A) The object is a wire, with density given in units of mass/length. Hence if we subdivide into pieces Δx units long, each will have nearly constant density and

Total mass ≐ ∑(2 + cos(πx))Δx

= ∫_{−2}^{3} 2 + cos(πx) dx

= (2x + \frac{1}{π} sin(πx))|_{−2}^{3}

= 6 + \frac{1}{π} sin(3π) − (−4 + \frac{1}{π} sin(−2π))

= 10 grams

B) To determine whether the wire balances at x = 1/2, we need to locate its center of mass. Since the density is oscillating, we cannot see immediately where this is, without computing \overline{x} explicitly (or analyzing the density function more closely). To compute the total moment, we set up the integral as usual, separate and then use parts on the second term (u = x, dv = cos(πx) dx, so du = dx and v = \frac{1}{π} sin(πx)):

∫_{−2}^{3} x(2 + cos(πx)) dx = ∫_{−2}^{3} 2x dx + ∫_{−2}^{3} x cos(πx) dx

= x^{2}|_{−2}^{3} + \frac{x}{π} sin(πx)|_{−2}^{3} − \frac{1}{π} ∫_{−2}^{3} sin(πx) dx

= 5 + 0 + \frac{1}{π^{2}} cos(πx)|_{−2}^{3}

= 5 − \frac{2}{π^{2}}

Hence

\overline{x} = \frac{5 − \frac{2}{π^{2}}}{10} = \frac{5π^{2} − 2}{10π^{2}} ≐ .48

Since the center of mass is slightly to the left of 1/2, the wire will tip to the left.