📝 Practice Problems

Solution:
Top: $y=4$, Bottom: $y=x^2$
$\text{Area} = \int_{-2}^2 (4 - x^2) dx = \left[4x - \frac{x^3}{3}\right]_{-2}^2 = (8 - \frac{8}{3}) - (-8 + \frac{8}{3}) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}$

Solution:
Top: $y=x$, Bottom: $y=x^3$
$\text{Area} = \int_0^1 (x - x^3) dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$

Solution:
Disk Method: $R(x) = \sqrt{x}$
$V = \pi \int_0^4 (\sqrt{x})^2 dx = \pi \int_0^4 x dx = \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi$

Solution:
Intersection: $x^2=x$ ⇒ $x=0,1$
Outer: $R(x)=x$, Inner: $r(x)=x^2$
$V = \pi \int_0^1 (x^2 - x^4) dx = \pi \left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1 = \pi(\frac{1}{3} - \frac{1}{5}) = \frac{2\pi}{15}$

Solution:
Shell Method: radius = $x$, height = $x^2$
$V = 2\pi \int_0^2 x \cdot x^2 dx = 2\pi \int_0^2 x^3 dx = 2\pi \left[\frac{x^4}{4}\right]_0^2 = 2\pi \cdot 4 = 8\pi$

Solution:
Intersection: $x=0,1$
Shell Method: radius = $x$, height = $x - x^2$
$V = 2\pi \int_0^1 x(x - x^2) dx = 2\pi \int_0^1 (x^2 - x^3) dx = 2\pi \left[\frac{x^3}{3} - \frac{x^4}{4}\right]_0^1 = 2\pi(\frac{1}{3} - \frac{1}{4}) = \frac{\pi}{6}$

Solution:
$y = \sqrt{x}$ ⇒ $x = y^2$, $y$ from $0$ to $2$
Disk Method: $R(y) = y^2$
$V = \pi \int_0^2 (y^2)^2 dy = \pi \int_0^2 y^4 dy = \pi \left[\frac{y^5}{5}\right]_0^2 = \frac{32\pi}{5}$

Solution:
$f_{\text{avg}} = \frac{1}{3-0} \int_0^3 x^2 dx = \frac{1}{3} \left[\frac{x^3}{3}\right]_0^3 = \frac{1}{3} \cdot 9 = 3$

Solution:
$f_{\text{avg}} = \frac{1}{\pi} \int_0^{\pi} \sin x dx = \frac{1}{\pi} [-\cos x]_0^{\pi} = \frac{1}{\pi} (1 + 1) = \frac{2}{\pi}$

Solution:
From Problem 8, $f_{\text{avg}} = 3$. So $c^2 = 3$ ⇒ $c = \sqrt{3} \approx 1.732$ (in $[0,3]$).

Solution:
$f'(x) = 3$, $L = \int_0^4 \sqrt{1+9} dx = \int_0^4 \sqrt{10} dx = 4\sqrt{10}$

Solution:
$f'(x) = 2x$, $L = \int_0^2 \sqrt{1 + 4x^2} dx$
$= \left[\frac{x}{2}\sqrt{1+4x^2} + \frac{1}{4}\ln\left(2x + \sqrt{1+4x^2}\right)\right]_0^2$
$= \frac{2}{2}\sqrt{17} + \frac{1}{4}\ln(4+\sqrt{17}) - 0 = \sqrt{17} + \frac{1}{4}\ln(4+\sqrt{17}) \approx 4.646$

Solution:
$f'(x) = \frac{1}{2\sqrt{x}}$, $(f'(x))^2 = \frac{1}{4x}$
$L = \int_0^4 \sqrt{1 + \frac{1}{4x}} dx$ (improper at $x=0$)

Solution:
Intersection: $y^2 = y+2$ ⇒ $y^2-y-2=0$ ⇒ $y=-1,2$
Right: $x = y+2$, Left: $x = y^2$
$\text{Area} = \int_{-1}^2 [(y+2) - y^2] dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2 = \frac{9}{2}$

Solution:
Intersection: $x^2=4$ ⇒ $x=\pm2$, $y$ from $0$ to $4$
Outer: $x = \sqrt{y}$, Inner: $x = 0$
$V = \pi \int_0^4 (\sqrt{y})^2 dy = \pi \int_0^4 y dy = \pi \left[\frac{y^2}{2}\right]_0^4 = 8\pi$

Solution:
Shell method around x‑axis: integrate with respect to $y$
$x = y^2$, radius = $y$, height = $y^2$
$V = 2\pi \int_0^2 y \cdot y^2 dy = 2\pi \int_0^2 y^3 dy = 2\pi \cdot 4 = 8\pi$

Solution:
$f_{\text{avg}} = \frac{1}{2} \int_0^2 e^x dx = \frac{1}{2} (e^2 - 1)$

Solution:
Radius = $4 - x^2$
$V = \pi \int_0^2 (4 - x^2)^2 dx = \pi \int_0^2 (16 - 8x^2 + x^4) dx = \pi \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_0^2 = \pi \left(32 - \frac{64}{3} + \frac{32}{5}\right) = \frac{256\pi}{15}$

Solution:
$f'(x) = \frac{1}{x}$, $(f'(x))^2 = \frac{1}{x^2}$
$L = \int_1^3 \sqrt{1 + \frac{1}{x^2}} dx = \int_1^3 \frac{\sqrt{x^2+1}}{x} dx$

Solution:
Use washers with respect to $y$: $x = \sqrt{y}$ and $x = -\sqrt{y}$
Outer radius = $2 - (-\sqrt{y}) = 2 + \sqrt{y}$
Inner radius = $2 - \sqrt{y}$
$V = \pi \int_0^4 [(2+\sqrt{y})^2 - (2-\sqrt{y})^2] dy = \pi \int_0^4 8\sqrt{y} dy = \pi \cdot 8 \cdot \frac{2}{3} [y^{3/2}]_0^4 = \frac{16\pi}{3} \cdot 8 = \frac{128\pi}{3}$