7.1, #21.

∫ (\frac{3}{t} − \frac{2}{t^{2}} ) dt = 3 ln |t| + 2t^{−1} + C

7.1, #26.

∫ (e^{x} + 5) dx = e^{x} + 5x + C

7.1, #43. The indefinite integral is 7.1, #14 on assignment 6.

∫_{1}^{2} \frac{1 + y^{2}}{y} dy = (ln y + y^{2}/2)|_{1}^{2} = ln 2 + 3/2 = 2.193

7.1, #54. The average value of sin t on the interval 0 ≤ t ≤ 2π is

\frac{1}{2π − 0} ∫_{0}^{2π} sin t dt = 0.

We can evaluate this integral using an antiderivative or by knowing that the areas above and below the x-axis are equal. This is becasue sin(t + π) = − sin t. The part of the graph above the x-axis balances the part below. This is a huristic argument that the average should be 0.

The average value of sin t on 0 ≤ t ≤ π is

\frac{1}{π} ∫_{0}^{π} sin t dt = \frac{1}{π} [− cos t]_{0}^{π} = \frac{1}{π} (−(−1) − (−1)) = \frac{2}{π} .

7.2, #2. Use the substitution u = x^{2}, du = 2x dx to get

∫ 2x cos(x^{2}) dx = ∫ cos u du = sin u + C = sin(x^{2}) + C

7.2, #5. Use the substitution u = sin x, du = cos x dx to get

∫ e^{sin x} cos x dx = ∫ e^{u} du = e^{u} + C = e^{sin x} + C

7.2, #6. Use the substitution u = x^{2} + 1, du = 2x dx to get

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