18.01 Solutions, Problem Set 5

• §5.2 #4.

Calculate d (\frac{x − 2}{x + 3}).

Solution:

d (\frac{x − 2}{x + 3}) = \frac{(x + 3)d(x − 2) − (x − 2)d(x + 3)}{(x + 3)^{2}} (quotient rule)

= \frac{[(x + 3) − (x − 2)] dx}{(x + 3)^{2}} = \frac{5}{(x + 3)^{2}} dx.

• §5.2 #9.

Calculate d(x^{2}\sqrt{3x + 2}).

Solution:

d(x^{2}\sqrt{3x + 2}) = x^{2}d(3x + 2)^{1/2} + (3x + 2)^{1/2}dx^{2}  (product rule)
= x^{2} · \frac{1}{2}(3x + 2)^{−1/2}d(3x + 2) + (3x + 2)^{1/2}2x dx  (power rule)

= \frac{3x^{2} dx}{2(3x + 2)^{1/2}} + \frac{4x(3x + 2) dx}{2(3x + 2)^{1/2}} = \frac{15x^{2} + 8x}{2(3x + 2)^{1/2}} dx.

• §5.2 #16.

Use differentials to approximate ^{4} \sqrt{83}. Use a calculator to find the value correct to six decimal values, and compare.

Solution: If y = x^{1/4}, then dy = \frac{1}{4}x^{−3/4}dx. Applying this with x = 81 and dx = 2, we get y = ^{4} \sqrt{81} = 3 and dy = \frac{1}{4}(81)^{−3/4} · 2 = 1/54. So

\sqrt{83} ≈ 3 + 1/54 = 163/54.

A calculator gives \sqrt{83} ≈ 3.0183495, compared to 163/54 ≈ 3.0185185.

• §5.3 Compute the following indefinite integrals.

#3. ∫ (x^{2} + x^{3} + x^{4}) dx. Solution:

∫ (x^{2} + x^{3} + x^{4}) dx = \frac{1}{3}x^{3} + \frac{1}{4}x^{4} + \frac{1}{5}x^{5} + C.

#14. ∫ x^{2}\sqrt{x} dx.

Solution:

∫ x^{2}\sqrt{x} dx = ∫ x^{5/2}dx = \frac{2}{7}x^{7/2} + C.

#22. ∫ (\frac{6}{x^{4}} + 2x^{3/2} − 5) dx.

Solution:

\begin{matrix}∫ (\frac{6}{x^{4}} + 2x^{3/2} − 5) dx = ∫ (6x^{−4} + 2x^{3/2} − 5) dx \\ = 6 · \frac{1}{−3}x^{−3} + 2 · \frac{2}{5}x^{5/2} − 5x + C = \frac{−2}{x^{3}} + \frac{4}{5}x^{5/2} − 5x + C. \end{matrix}

#43. ∫ \frac{(1 + \sqrt{x})^{1/4}}{\sqrt{x}} dx, using the substitution u = 1 + \sqrt{x}.

Solution: If u = 1 + \sqrt{x}, then du = \frac{1}{2\sqrt{x}} dx, so

\begin{aligned}∫ \frac{2(1 + \sqrt{x})^{1/4}}{2\sqrt{x}} dx & = ∫ 2u^{1/4} du \\ = \frac{8}{5}u^{5/4} + C & = \frac{8}{5}(1 + \sqrt{x})^{5/4} + C. \end{aligned}

#56. ∫ x^{2} ^{3} \sqrt{x^{3} + 1} dx, using a substitution of your own devising. Solution: Let u = x^{3} + 1, so du = 3x^{2} dx and

\begin{aligned}∫ x^{2} ^{3} \sqrt{x^{3} + 1} dx & = ∫ \frac{1}{3}u^{1/3}du \\ = \frac{1}{3} · \frac{3}{4}u^{4/3} + C & = \frac{1}{4}(x^{3} + 1)^{4/3} + C. \end{aligned}

• p. 288 Evaluate the following integrals.

#7. ∫ e^{−3x} dx. Solution: Let u = −3x, so du = −3dx and

∫ e^{−3x}dx = ∫ −\frac{1}{3}e^{u} du = −\frac{1}{3}e^{u} + C = −\frac{1}{3}e^{−3x} + C.

#10. ∫ \frac{4 dx}{\sqrt{e^{x}}}.

Solution: We write \sqrt{e^{x}} = (e^{x})^{1/2} = e^{x/2}, so \frac{1}{\sqrt{e^{x}}} = e^{−x/2}. The substitution u = −x/2 gives

∫ 4e^{−x/2}dx = ∫ −8e^{u}du == −8e^{u} + C = −8e^{−x/2} + C = \frac{−8}{\sqrt{e^{x}}} + C.

#25d) ∫_{0}^{3} \frac{x dx}{x^{2} + 1}.

Solution: Let u = x^{2} + 1, so du = 2x dx and

∫ \frac{x dx}{x^{2} + 1} = ∫ \frac{du}{2u} = \frac{1}{2} ln |u| + C = \frac{1}{2} ln(x^{2} + 1) + C.

Evaluating the definite integral then gives

∫_{0}^{3} \frac{x dx}{x^{2} + 1} = \frac{1}{2} ln 10 − \frac{1}{2} ln 1 = \frac{1}{2} ln 10.

#25k) ∫ \frac{e^{x} dx}{e^{x} + 1}.

Solution: Let u = e^{x} + 1, so du = e^{x}dx and

∫ \frac{e^{x} dx}{e^{x} + 1} = ∫ \frac{du}{u} = ln |u| + C = ln(e^{x} + 1) + C.

#25m) ∫ \frac{(ln x)^{2}dx}{x}.

Solution: Let u = ln x, so du = \frac{1}{x} dx and

∫ \frac{(ln x)^{2}dx}{x} = ∫ u^{2}du = \frac{1}{3}u^{3} + C = \frac{1}{3}(ln x)^{3} + C.

• p. 331 Evaluate the indefinite integrals. #37. ∫ cos(3x) dx.

Solution: Let u = 3x, so du = 3dx and

∫ cos(3x) dx = ∫ \frac{1}{3} cos u du = \frac{1}{3} sin u + C = \frac{1}{3} sin(3x) + C.

#40. ∫ cos^{2}(7x) sin(7x) dx.

Solution: Let u = 7x, so du = 7dx and

∫ cos^{2}(7x) sin(7x) dx = ∫ \frac{1}{7} cos^{2} u sin u du.

Now substitute again, letting v = cos u, so dv = − sin u du and

∫ \frac{1}{7} cos^{2} u sin u dx = ∫ \frac{−1}{7}v^{2}dv = −\frac{1}{21}v^{3} + C.

Since v = cos u = cos(7x), this gives ∫ cos^{2}(7x) sin(7x) dx = −\frac{1}{21} cos^{3}(7x) + C.

#45. ∫ x^{2} cos(x^{3}) dx.

Solution: Let u = x^{3}, so du = 3x^{2}dx and

∫ x^{2} cos(x^{3}) dx = ∫ \frac{1}{3} cos u du = \frac{1}{3} sin u + C = \frac{1}{3} sin(x^{3}) + C.

• §6.6 #2. Use integration to find the area of the triangle bounded by the axes and the line 3x + 2y = 6. Check your answer using elementary geometry.

Solution: The line is the graph of y = −\frac{3}{2}x+3. We want the area under the graph, between x = 0 and x = 2 (the x-intercept). This is given by the integral

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

We are computing the area of a triangle with base 2 and height 3. The formula A = \frac{1}{2}bh from elementary geometry gives 3 as well.

• §6.6 #7. The curve y = x^{3} − 9x has one arch above the x-axis. Find the area of the region under the arch.

Solution: The arch is between the x-intercepts at x = −3 and x = 0, so the area of the region is

∫_{−3}^{0}(x^{3} − 9x) dx = (\frac{1}{4}x^{4} − \frac{9}{2}x^{2})]_{−3}^{0} = 0 − (\frac{81}{4} − \frac{81}{2}) = \frac{81}{4}.

• §6.6 #15.

Find the area bounded by the curve y = 2x + \frac{1}{x^{2}} , the x-axis and the lines x = 1 and x = 3. Solution: Note that if x > 0, then y > 0, so the region is above the x-axis. The area is

∫_{1}^{3} (2x + \frac{1}{x^{2}} ) dx = (x^{2} − \frac{1}{x})]_{1}^{3} = (9 − \frac{1}{3}) − (1 − 1) = \frac{26}{3}.

• §6.6 # 22.

If n is positive, then

∫_{−1}^{1} x^{n} dx = \frac{x^{n+1}}{n + 1}]_{−1}^{1} .

Why is this calculation incorrect if n is a negative number ≠ −1? Solution: The calculation uses the Fundamental Theorem of Calculus, which applies to ∫_{a}^{b} f(x) dx when f(x) is continuous on [a, b]. If n is negative, then f(x) = x^{n} is not continuous on the interval [−1, 1]. It is not even defined at x = 0, so we haven’t even defined the integral ∫_{−1}^{1} x^{n} dx.

• §6.6 Find the value of each definite integral. #26. ∫_{1}^{4} \sqrt{x} dx.

Solution:

∫_{1}^{4} x^{1/2}dx = \frac{2}{3}x^{3/2}]_{1}^{4} = \frac{2}{3}(8 − 1) = \frac{14}{3}.

#31. ∫_{0}^{1}(x − x^{2}) dx.

Solution:

∫_{0}^{1}(x − x^{2}) dx = (\frac{1}{2}x^{2} − \frac{1}{3}x^{3})]_{0}^{1} = \frac{1}{2} − \frac{1}{3} = \frac{1}{6}.

#38. ∫_{1}^{2} (x + \frac{1}{x})^{2} dx.

Solution:

\begin{aligned}∫_{1}^{2} (x + \frac{1}{x})^{2} dx & = ∫_{1}^{2}(x^{2} + 2 + x^{−2}) dx = (x^{3}/3 + 2x − x^{−1})]_{1}^{2} \\ & = (\frac{8}{3} + 4 − \frac{1}{2}) − (\frac{1}{3} + 2 − 1) = \frac{29}{6}. \end{aligned}

#40. ∫_{π/6}^{π/4} sin t dt.

Solution:

∫_{π/6}^{π/4} sin t dt = − cos t]_{π/6}^{π/4} = cos(π/6)−cos(π/4) = \sqrt{3}/2−\sqrt{2}/2 = \frac{\sqrt{3} − \sqrt{2}}{2}.

• §6.7 #3. Find the area bounded by y^{2} = x^{3} and x = 4.

Solution: This is twice the area under the graph of y = x^{3/2} between x = 0 and x = 4, so the area is given by

2 ∫_{0}^{4} x^{3/2}dx = \frac{4}{5}x^{5/2}]_{0}^{4} = \frac{4}{5} · 4^{5/2} = \frac{128}{5}.

• §6.7 #9. If f(x) is an odd function, that is, if f(−x) = −f(x), show geometrically or otherwise that

∫_{−a}^{a} f(x) dx = 0.

Solution: By the interval addition property of definite integrals,

∫_{−a}^{a} f(x) dx = ∫_{−a}^{0} f(x) dx + ∫_{0}^{a} f(x) dx.

Since f(x) is odd, f(x) = −f(−x), so the first integral in the sum is the

negative of ∫_{−a}^{0} f(−x) dx. The graph of y = f(−x) on the interval [−a, 0]

is the mirror image in the y-axis of the graph of y = f(x) on [0, a], so the two integrals

∫_{−a}^{0} f(−x) dx, ∫_{0}^{a} f(x) dx

coincide and it follows that ∫_{−a}^{a} f(x) dx = 0.

• §6.7 #11. The graph of y = x^{2}, x ≥ 0, can be considered to be the graph of x = \sqrt{y}, y ≥ 0. Show by geometry that this implies the validity of the equation

∫_{0}^{a} x^{2} dx + ∫_{0}^{a^{2}} \sqrt{y} dy = a^{3}.

Solution: The first integral measures the area of the part of the rectangle underneath the curve, the second measures the area of the part above the curve. Their sum is therefore the area of the rectangle, which is a^{3}. Calculating the integrals gives

\begin{aligned}∫_{0}^{a} x^{2} dx & = \frac{1}{3}x^{3}]_{0}^{a} & = \frac{a^{3}}{3} \\ ∫_{0}^{a^{2}} y^{1/2} dy & = \frac{2}{3}a^{3/2}]_{0}^{a^{2}} & = \frac{2a^{3}}{3} . \end{aligned}

So this also shows the sum is a^{3}.

• §6.7 #13. Use the known area of a circle to find the value of the integral

∫_{−a}^{a} \sqrt{a^{2} − x^{2}} dx.

Solution: The curve y = \sqrt{a^{2} − x^{2}} (with −a ≤ x ≤ a) is the top half (y ≥ 0) of the circle x^{2} + y^{2} = a^{2} (with radius a, center (0, 0)). So the value of the integral is half the area of the circle, so

∫_{−a}^{a} \sqrt{a^{2} − x^{2}} dx = \frac{1}{2}πa^{2}.

• §6.7 #16

Compute the indicated derivative

d) \frac{d}{dx} ∫_{x}^{1} \frac{dt}{1 + t^{4}} .

Solution: Since

∫_{x}^{1} \frac{dt}{1 + t^{4}} = − ∫_{1}^{x} \frac{dt}{1 + t^{4}} ,

we have

\frac{d}{dx} ∫_{x}^{1} \frac{dt}{1 + t^{4}} = \frac{−1}{1 + x^{4}} .

(by equation (13) of §6.7, or the “Second Fundamental Theorem of Calculus”).

e) \frac{d}{dx} ∫_{1}^{x^{2}} \frac{dt}{\sqrt{t + \sqrt{t + 1}}}.

Solution: Note that the upper limit of integration is x^{2}, not x. To apply equation (13), we use the chain rule with u = x^{2}. This gives

\frac{d}{dx} ∫_{1}^{x^{2}} \frac{dt}{\sqrt{t + \sqrt{t + 1}}} = \frac{d}{du} ∫_{1}^{u} \frac{dt}{\sqrt{t + \sqrt{t + 1}}} \frac{du}{dx}

= \frac{1}{\sqrt{u + \sqrt{u + 1}}} · 2x

= \frac{2x}{\sqrt{x^{2} + \sqrt{x^{2} + 1}}}.

• §10.9 #2. Clearly,

∫_{0}^{π} sin x dx = 2.

Calculate the value of this integral approximately using Simpson’s rule with n = 4.

Solution: Divding the interval [0, π] into n = 4 equal subintervals, the endpoints are

\begin{matrix}x_{0} = 0, & x_{1} = π/4, & x_{2} = π/2, & x_{3} = 3π/4, & x_{4} = π; so \\ y_{0} = 0, & y_{1} = \sqrt{2}/2, & y_{2} = 1, & y_{3} = \sqrt{2}/2, & y_{4} = 0. \end{matrix}

Simpson’s rule gives

∫_{0}^{π} sin x dx ≈ \frac{π}{12}(0+4·\sqrt{2}/2+2·1+4·\sqrt{2}/2+0) = (1+2\sqrt{2})\frac{π}{6} ≈ 2.00456.

• §10.9 #5. The exact value of

∫_{0}^{2} e^{−x^{2}}dx

is not known, but to 10 decimal places it is 0.8820813908. Calculate this integral approximately using Simpson’s rule with n = 4.

Solution: We have

\begin{matrix}x_{0} = 0, & x_{1} = 1/2, & x_{2} = 1, & x_{3} = 3/2, & x_{4} = 2; so \\ y_{0} = 1, & y_{1} = e^{−1/4}, & y_{2} = e^{−1}, & y_{3} = e^{−9/4} & y_{4} = e^{−4}. \end{matrix}

Simpson’s rule gives

∫_{0}^{2} e^{−x^{2}}dx ≈ \frac{1}{6}(1 + 4 · 0.7788 + 2 · 0.3679 + 4 · 0.1054 + 0.0183) ≈ 0.8818.

• §10.9 #9 The width, in feet, at equally spaced points along the fairway of a hole on a golf course is given in Fig. 10.7. The management wishes to estimate the number of square yards of the fairway as a basis for deciding how long it should take a groundskeeper to mow it. Use Simpson’s rule to provide such an estimate.

Solution: Applying Simpson’s rule with n = 12 gives the following approximation of the area:

\frac{1200}{36}(0+4·180+2·190+4·175+2·170+4·185+2·195+4·205+2·210+4·200+2·195+4·170+0)

≈ 212,667 square feet, or 23,630 square yards.