7.5, #46. The integrand is an improper rational function. Long division gives

\frac{x^{2}}{x^{2} + 6x + 13} = 1 − \frac{6x + 13}{x^{2} + 6x + 13}.

Completing the square gives x^{2} + 6x + 13 = (x + 3)^{2} + 2^{2}, so set w = x + 3 and dw = dx. Then

\frac{6x + 13}{x^{2} + 6x + 13} = \frac{6w − 5}{w^{2} + 2^{2}} ,

so formula 25 applies with a = 2, b = 6, c = −5, giving

∫ \frac{x^{2}}{x^{2} + 6x + 13} dx = x − 3 ln |(x + 3)^{2} + 2^{2}| + \frac{5}{2} arctan \frac{x + 3}{2} + C.

7.5, #50. The integrand is an improper rational function. Long division gives

\frac{t^{2} + 1}{t^{2} − 1} = 1 + \frac{2}{t^{2} − 1}.

Since t^{2} − 1 = (t − 1)(t + 1), formula 26 with a = 1, b = −1 gives

∫ \frac{t^{2} + 1}{t^{2} − 1} dt = t + ln |t − 1| − ln |t + 1| + C.

7.5, #51. Formula 24 with a = 2 gives

∫_{0}^{2} \frac{dx}{4 + x^{2}} = \frac{1}{2} arctan \frac{x}{2}|_{0}^{2} = \frac{1}{2}(arctan 1 − arctan 0) = \frac{π}{8} = 0.3927.

The left and right Riemann sums with n = 100 are approximately 0.3939 and 0.3915. Since the integrand is decreasing on the interval [0, 2], its value is between these two estimates.

7.5, #53. Since x^{2} + 2x + 5 = (x + 1)^{2} + 2^{2}, formula 24 with a = 2 gives

∫_{0}^{1} \frac{dx}{x^{2} + 2x + 5} = \frac{1}{2} arctan \frac{x + 1}{2}|_{0}^{1} = \frac{1}{2}(arctan 1 − arctan \frac{1}{2}) = 0.1609.

The left and right Riemann sums with n = 100 are approximately 0.1613 and 0.1605. Since the integrand is decreasing on the interval [0, 1], its value is between these two estimates. 7.5, #64. (a) The average voltage over a second is

\frac{1}{1 − 0} ∫_{0}^{1} V_{0} cos(120πt) dt = \frac{V_{0}}{120π} sin(120πt)|_{0}^{1} = 0.

(b) The average of V^{2} over a second is

\frac{1}{1 − 0} ∫_{0}^{1} V_{0}^{2} cos^{2}(120πt) dt = \frac{V_{0}^{2}}{120π} ∫_{0}^{120π} cos^{2} u du

= \frac{V_{0}^{2}}{240π} (cos u sin u + u)|_{0}^{120π} = \frac{1}{2}V_{0}^{2}

using the substitution u = 120πt and du = 120π dt. Thus \overline{V} = V_{0}/\sqrt{2}.

(c) If \overline{V} = 110, then V_{0} = 110\sqrt{2} = 155.56 volts.

7.6, #1.

n = 1 n = 2 n = 4
LEFT  40.0000 40.7846 41.7116
RIGHT 51.2250 46.3971 44.5179
TRAP 45.6125 43.5909 43.1147
MID  41.5692 42.6386 42.8795

7.6, #3.

n 10 100 1000
LEFT  5.4711 5.8116 5.8464
RIGHT 6.2443 5.8890 5.8541
TRAP 5.8577 5.8503 5.8502
MID  5.8465 5.8502 5.8502

e^{x}\sqrt{x} is increasing and concave-up on [1, 2] since

(e^{x}\sqrt{x})′ = e^{x} (\sqrt{x} + \frac{1}{2\sqrt{x}}) , (e^{x}\sqrt{x})′′ = e^{x} (\sqrt{x} + \frac{1}{\sqrt{x}} − \frac{1}{4\sqrt{x^{3}}} )

are positive on [1, 2]. (\sqrt{x} ≥ 1 and 1/(4\sqrt{x^{3}}) ≤ 1/4 for 1 ≤ x ≤ 2.) So LEFT and MID underestimate the integral, RIGHT and TRAP overestimate the integral.

7.6, #4.

n 10 100 1000
LEFT  3.0132 2.9948 2.9930
RIGHT 2.9711 2.9906 2.9925
TRAP 2.9922 2.9927 2.9927
MID  2.9930 2.9927 2.9927

\sqrt{3 + cos θ} is decreasing and concave-down on [0, π/2] since

(\sqrt{3 + cos θ})′ = \frac{− sin θ}{2\sqrt{3 + cos θ}} ,
(\sqrt{3 + cos θ})′′ = \frac{2(3 + cos θ)^{1/2}(− cos θ) + sin θ(3 + cos θ)^{−1/2}(− sin θ)}{4(3 + cos θ)}

are negative on [0, π/2]. (The second derivative is messy, but its sign is easy to determine. The numerator is always negative since square roots are non-negative; the denominator is positive for 0 ≤ θ ≤ π/2.) So LEFT and MID overestimate the integral, RIGHT and TRAP underestimate the integral.