
Set up the integral to find the volume of this region enclosed by y=\sqrt{-x} and x=-4 and x=0, rotated about the x-axis using the disk method.
V=\int_-4^0 \ \pi (\sqrt{-x})^2 \ dx
Use Integration by Parts to evaluate:
\int_2^3 \ \ln(x^2) \ dx
Combine the logarithmic terms in your answer!
ln(729/16)-6
Solve the trigonometric integral:
\int \ sin(-10x)cos(10x) \ dx
\frac{cos(-20x)}{40}+C
or
\frac{cos(20x)}{40}+C

Set up the integral to find the volume of this region, rotated about the y-axis using the shell method.
V = \int_1^2 2 \pi x (\frac{x}{2}-\frac{2}{x}) dx
Find the arc length of a function whose derivative is
\frac{dy}{dx}=\sqrt{x^2+4x+3}
over the interval [0,2].
Solve the trigonometric integral:
\int \ \sin^5(2t)\cos^2(2t)dt
- \frac1 6 cos^3(2t)+ \frac1 5 cos^5(2t) -\frac1 14 cos^7(2t) +C
**DAILY DOUBLE**
A region is enclosed by the equations
y=x^2+2 and y = -2x+2
rotated about the line x = 2. Set up the integrals to solve for the volume using the washer AND shell method.
Washer:
V=\int_2^6 \ pi(2-(1-\frac{y}{2}))^2 - pi(2-(-sqrt{y-2}))^2 \ dy
Shell:
V = \int_-2^0 \ 2 pi (2-x) ((-2x+2)-(x^2+2)) \ dx
Use Integration by Parts to evaluate:
\int \ x^2 e^{8x} dx
frac{x^2e^{8x}}{8} - frac{xe^{8x}}{32} + frac{e^{8x}}{256}+C
Use trigonometric substitutions to solve the integral:
\int \frac{dx}{sqrt{x^6-x^4}}
\frac{\sqrt{x^2-1}}{x}+C