Average Value of a Function

$f(x) = x^2 - 2x + 3$ on $[0, 3]$.

$\Big[\tfrac{x^3}{3} - x^2 + 3x\Big]_0^3 = 9 - 9 + 9 = 9$.

$\bar f = \dfrac{9}{3 - 0} = 3$.

$f(x) = \sin x\cos^2 x$ on $\left[0, \tfrac{\pi}{2}\right]$.

$u = \cos x$, $du = -\sin x\,dx$: $\displaystyle\int_0^{\pi/2}\sin x\cos^2 x\,dx = \int_0^1 u^2\,du = \tfrac{1}{3}$.

$\bar f = \dfrac{1/3}{\pi/2} = \dfrac{2}{3\pi}$.

$f(x) = xe^{-x}$ on $[0, 2]$.

Parts with $u = x$, $dv = e^{-x}\,dx$: $\Big[-xe^{-x} - e^{-x}\Big]_0^2 = 1 - 3e^{-2}$.

$\bar f = \dfrac{1 - 3e^{-2}}{2}$.

$T(t)$ on $[0, 12]$ hours.

$\Big[12t - \frac{48}{\pi}\cos\left(\frac{\pi t}{12}\right)\Big]_0^{12} = \left(144 + \frac{48}{\pi}\right) - \left(-\frac{48}{\pi}\right) = 144 + \frac{96}{\pi}$.

$\bar T = \dfrac{1}{12}\left(144 + \dfrac{96}{\pi}\right) = 12 + \dfrac{8}{\pi}$.

$f(x)^2 = (x^2 - 1)^2 = x^4 - 2x^2 + 1$.

$\dfrac{1}{2}\displaystyle\int_{-1}^{1}(x^4 - 2x^2 + 1)\,dx = \dfrac{1}{2}\left(\dfrac{2}{5} - \dfrac{4}{3} + 2\right) = \dfrac{8}{15}$.

$\sqrt{\dfrac{8}{15}} \approx 0.73$.

$V(t)^2 = 170^2\sin^2(120\pi t) = \dfrac{170^2}{2}\big(1 - \cos(240\pi t)\big)$.

Over $\left[0, \frac{1}{60}\right]$ the cosine runs through whole periods and integrates to $0$, so the average of $V^2$ is $\dfrac{170^2}{2}$.

$\sqrt{\dfrac{170^2}{2}} = \dfrac{170}{\sqrt{2}} \approx 120$ V.

$f(x)^2 = \dfrac{1}{x^2}$.

$\dfrac{1}{3}\displaystyle\int_1^4 \frac{dx}{x^2} = \dfrac{1}{3}\left(1 - \dfrac{1}{4}\right) = \dfrac{1}{4}$.

$\sqrt{\tfrac{1}{4}} = \tfrac{1}{2}$.

Total $= \bar f\,(b - a)$.

$\bar f = 40$ L/min and $b - a = 15$ min.

Total $= 40 \times 15 = 600$ L.

$5 \cdot (2 - 0) = \displaystyle\int_0^2 (kx^2 + 1)\,dx$.

$\displaystyle\int_0^2 (kx^2 + 1)\,dx = \frac{8k}{3} + 2$, so $10 = \frac{8k}{3} + 2$.

$\frac{8k}{3} = 8$, so $k = 3$.

$4 \cdot (b - 0) = \displaystyle\int_0^b \sqrt{x}\,dx$.

$\displaystyle\int_0^b \sqrt{x}\,dx = \tfrac{2}{3}b^{3/2}$, so $4b = \tfrac{2}{3}b^{3/2}$.

Divide by $b > 0$: $\sqrt{b} = 6$, so $b = 36$.

$\bar f \cdot (6 - 0) = \displaystyle\int_0^6 (\text{rate})\,dt$.

Trapezoidal rule, $\Delta t = 2$: $\displaystyle\int_0^6 (\text{rate})\,dt \approx \tfrac{2}{2}\big(10 + 2(14) + 2(12) + 8\big) = 70$ m$^3$.

$\bar f \approx \dfrac{70}{6} = \dfrac{35}{3} \approx 11.7$ m$^3$/h.