MATH 1A: CALCULUS HOMEWORK #13

JOHN VOIGHT

§5.3: The Fundamental Theorem of Calculus

Problem 4. For (a), we note that g(−3) = 0 (the integral is empty) and g(3) = 0 as well (the areas above and below the axis are equal).

For (b), we estimate that

g(−2) = ∫_{−3}^{−2} f(t) dt

is about the area of a triangle with base 1 and height 2, so g(−2) ≈ 1. For g(−1) we add approximately 2\frac{1}{2} (counting boxes) so g(−1) = 1 + 2\frac{1}{2} ≈ 3\frac{1}{2}. For g(0) we add finally a triangle which has area 3/2, so g(0) = 3\frac{1}{2} + 1\frac{1}{2} ≈ 6.

For (c), we note that g is increasing as long as we are adding more area, i.e. when the function f is positive. Therefore g is increasing on the interval (−3, 0) and is decreasing on (0, 3).

For (d), we note that g has a maximum value at x = 0 by (c) and the first derivative test.

For (e), we have the following graph, using parts (a)–(d):

For (f), note that the graph of g′ and the graph of f are the same by FTC.

Problem 8. Let f(t) = ln t so g(x) = ∫_{1}^{x} f(t) dt. Then by FTC we have g′(x) = f(x) = ln x.

Problem 10. Let f(x) = 1/(x+x^{2}) so g(u) = ∫_{3}^{u} f(x) dx. Then by FTC we have g′(u) = f(u) = 1/(u+u^{2}).

Problem 16. Let f(t) = (t + sin t) and h(x) = cos x. Then by FTC we have

y′ = f(h(x))h′(x) = (cos x + sin(cos x)) (− sin x) = − sin x(cos x + sin(cos x)).

Problem 22.

∫_{0}^{4}(1 + 3y − y^{2}) dy = (y + 3\frac{y^{2}}{2} − \frac{y^{3}}{3} )_{0}^{4} = 4 + 3 · 8 − \frac{64}{3} − 0 = \frac{20}{3}.

Date: April 27, 2004.

§5.3: 4, 8, 10, 16, 22, 26, 30, 32, 40, 42, 50, 52, 62, 66; §5.4: 2, 4, 22, 28, 38, 40, 44, 48, 54, 56, 58.

Problem 26. The integral does not exist because the function f(x) = x^{−5} has a vertical asymptote at x = 0 and hence is discontinuous on the interval [−2, 3].

Problem 30.

∫_{1}^{4} x^{−1/2} dx = 2x^{1/2}|_{1}^{4} = 2(2 − 1) = 2.

Problem 32.

∫_{0}^{1}(3 + x^{3/2}) dx = (3x + \frac{2}{5}x^{5/2})_{0}^{1} = 3 + \frac{2}{5} − 0 = \frac{17}{5}.

Problem 40.

∫_{1}^{2}(4u^{−3} + u^{−1}) du = (\frac{4}{−2}u^{−2} + ln |u|)_{1}^{2} = (−\frac{2}{u^{2}} + ln u)_{1}^{2} = −\frac{1}{2} + ln 2 + 2 − 0 = \frac{3}{2} + ln 2.

Problem 42.

∫_{−π}^{π} f(x) dx = ∫_{−π}^{0} x dx + ∫_{0}^{π} sin x dx = \frac{x^{2}}{2} |_{−π}^{0} − cos x|_{0}^{π} = 0 − \frac{π^{2}}{2} − (cos π − cos 0) = −\frac{π^{2}}{2} + 2.

Problem 50. Let h_{1}(x) = tan x and h_{2}(x) = x^{2}, f(t) = 1/\sqrt{2 + t^{4}}. Then by the FTC,

g′(x) = f(h_{2}(x))h′_{2}(x) − f(h_{1}(x))h′_{1}(x) = \frac{1}{\sqrt{2 + x^{8}}} (2x) − \frac{1}{\sqrt{2 + tan^{4} x}}(sec^{2} x) = \frac{2x}{\sqrt{2 + x^{8}}} − \frac{sec^{2} x}{\sqrt{2 + tan^{4} x}}.

Problem 52. Let h_{1}(x) = cos x and h_{2}(x) = 5x, f(u) = cos(u^{2}). Then by the FTC,

g′(x) = f(h_{2}(x))h′_{2}(x)−f(h_{1}(x))h′_{1}(x) = cos((5x)^{2})(5)−cos(cos^{2} x)(− sin x) = 5 cos(25x^{2})+sin x cos(cos^{2} x).

Problem 62. If we are to apply the Riemann sum, we break up the unit interval [0, 1] into n pieces, each of length 1/n. We have the formula

∫_{0}^{1} f(x) dx = lim_{n→∞} ∑_{i=1}^{n} f(x_{i})Δx.

It looks like we should take f(x) = \sqrt{x} (and the right endpoint), for then we would have

lim_{n→∞} ∑_{i=1}^{n} \sqrt{\frac{i}{n}} · \frac{1}{n}

which is what is given. Therefore the sum is

∫_{0}^{1} \sqrt{x} dx = \frac{2}{3}x^{3/2}|_{0}^{1} = \frac{2}{3}.

Problem 66. For (a), we break up the real line as the function is defined. For x < 0, g(x) = ∫_{0}^{x} f(t) dt = ∫_{0}^{x} 0 dt = 0. If 0 ≤ x ≤ 1, then

g(x) = ∫_{0}^{x} f(t) dt = ∫_{0}^{x} t dt = \frac{t^{2}}{2} |_{0}^{x} = \frac{x^{2}}{2} .

If 1 < x ≤ 2, then

g(x) = ∫_{0}^{x} f(t) dt = ∫_{0}^{1} t dt + ∫_{1}^{x}(2 − t) dt = g(1) + (2t − \frac{t^{2}}{2} )_{1}^{x} = \frac{1}{2} + 2x − \frac{x^{2}}{2} − 2 + \frac{1}{2} = −\frac{x^{2}}{2} + 2x − 1.

If x > 2, then

g(x) = ∫_{0}^{x} f(t) dt = g(2) + ∫_{2}^{x} 0 dt = −2 + 4 − 1 = 1.

Therefore

g(x) = \begin{cases}0, & x < 0 \\ \frac{x^{2}}{2} , & 0 ≤ x ≤ 1 \\ −\frac{x^{2}}{2} + 2x − 1, & 1 < x ≤ 2 \\ 1, & x > 2. \end{cases}

For (b), we have the following graphs of f and g, respectively:

Finally, for (c), we note that f is differentiable everywhere except at the corners, i.e. x ≠ 0, 1, 2. However, g is differentiable everywhere, since g′(x) = f(x) by the FTC!

5.4: Indefinite Integrals and the Net Change Theorem

Problem 2.

(x sin x + cos x + C)′ = sin x + x cos x − sin x = x cos x.

Problem 4. Remembering that a is a constant and we are differentiating with respect to x, we have

(−\frac{(x^{2} + a^{2})^{1/2}}{a^{2}x} + C)′ = −\frac{a^{2}x(1/2)(x^{2} + a^{2})^{−1/2}(2x) − (x^{2} + a^{2})^{1/2}(a^{2})}{a^{4}x^{2}} = −\frac{a^{2}x^{2}/\sqrt{x^{2} + a^{2}} − a^{2}\sqrt{x^{2} + a^{2}}}{a^{4}x^{2}} .

Now cancel a^{2} from the top and bottom and simplify the numerator by multiplying top and bottom by \sqrt{x^{2} + a^{2}}:

−\frac{x^{2} − (x^{2} + a^{2})}{a^{2}x^{2}\sqrt{x^{2} + a^{2}}} = \frac{a^{2}}{a^{2}x^{2}\sqrt{x^{2} + a^{2}}} = \frac{1}{x^{2}\sqrt{x^{2} + a^{2}}} .

Problem 22.

∫_{0}^{4}(2v + 5)(3v − 1) dv = ∫_{0}^{4}(6v^{2} + 13v − 5) dv = (2v^{3} + 13\frac{v^{2}}{2} − 5v)|_{0}^{4} = 128 + 104 − 20 = 212.

Problem 28.

∫_{0}^{5}(2e^{x} + 4 cos x) dx = (2e^{x} + 4 sin x)|_{0}^{5} = 2e^{5} + 4 sin 5 − 2 − 0 = 2e^{5} + 4 sin 5 − 2.

Problem 38.

∫_{4}^{9}(\sqrt{x}+1/\sqrt{x})^{2} dx = ∫_{4}^{9} (x + 2 + \frac{1}{x}) dx = (\frac{x^{2}}{2} + 2x + ln |x|)_{4}^{9} = \frac{81}{2} +18+ln 9−(8+8+ln 4) = \frac{85}{2} +ln \frac{9}{4}.

Problem 40. Break up the integral to get rid of the absolute value. Note that sin x ≥ 0 on this interval from [0, π] and sin x ≤ 0 on [π, 3π/2]. Therefore

∫_{0}^{3π/2} | sin x| dx = ∫_{0}^{π} sin x dx+∫_{π}^{3π/2} − sin x dx = (− cos x)_{0}^{π}+(cos x)_{π}^{3π/2} = (1−(−1))+(0−(−1)) = 2+1 = 3.

Problem 44. Since y = \sqrt[4]{x}, we have x = y^{4}. We integrate x = y^{4} from y = 0 to y = 1, so therefore

A = ∫_{0}^{1} y^{4} dy = \frac{y^{5}}{5} |_{0}^{1} = \frac{1}{5}.

Problem 48. By the Net Change Theorem, ∫_{0}^{15} n′(t) dt = n(15) − n(0) = n(15) − 100. This represents the increase in the bee population in 15 weeks. So 100 + ∫_{0}^{15} n′(t) dt = n(15) represents the total bee population after 15 weeks.

Problem 54. For (a), we have displacement given by

∫_{1}^{6} v(t) dt = ∫_{1}^{6}(t^{2} − 2t − 8) dt = (\frac{t^{3}}{3} − t^{2} − 8t)_{1}^{6} = (72 − 36 − 48) − (1/3 − 1 − 8) = −\frac{10}{3} m.

For (b), displacement is the absolute value of the distance travelled:

∫_{1}^{6} |v(t)| dt = ∫_{1}^{6} |t^{2} − 2t − 8| dt.

Now t^{2} − 2t − 8 = (t − 4)(t + 2) and this is ≥ 0 when t > 4 or t < −2 and ≤ 0 for t in the interval [−2, 4]. Therefore

∫_{1}^{6} |t^{2} − 2t − 8| dt = ∫_{1}^{4} −(t^{2} − 2t − 8) dt + ∫_{4}^{6}(t^{2} − 2t − 8) dt = − (\frac{t^{3}}{3} − t^{2} − 8t)_{1}^{4} + (\frac{t^{3}}{3} − t^{2} − 8t)_{4}^{6}

= (−64/3 + 16 + 32) − (−1/3 + 1 + 8) + (72 − 36 − 48) − (64/3 − 16 − 32) = \frac{98}{3} m.

Problem 56. For (a), v′(t) = a(t) = 2t + 3, so v(t) = t^{2} + 3t + C. Since v(0) = C = −4, v(t) = t^{2} + 3t − 4. For (b), we have the distance travelled (as in 54(b))

∫_{0}^{3} |t^{2} + 3t − 4| dt = ∫_{0}^{3} |(t + 4)(t − 1)| dt = ∫_{0}^{1} −(t^{2} + 3t − 4) dt + ∫_{1}^{3}(t^{2} + 3t − 4) dt

= − (\frac{t^{3}}{3} + 3\frac{t^{2}}{2} − 4t)_{0}^{1} + (\frac{t^{3}}{3} + 3\frac{t^{2}}{2} − 4t)_{1}^{3}

= (−1/3 − 3/2 + 4) + (9 + 27/2 − 12) − (1/3 + 3/2 − 4) = \frac{89}{6} m.

Problem 58. By the Net Change Theorem, the amount of water that flows from the tank is

∫_{0}^{10} r(t) dt = ∫_{0}^{10}(200 − 4t) dt = (200t − 2t^{2})_{0}^{10} = 2000 − 200 = 1800 liters.

Yeehaw!