# Thread: Help with Limit and Integration

1. ## Help with Limit and Integration

1. Find the limit: lim x->0

(cos 3x)^1/x (Hint: use logarithmic method)

2. let f(x)=(5x^2 + 1)/((x-1)(x+1)^2) Find a,b,c such that f(x)=(a/(x-1))+(b/(x+1))+(c/(x+1)^2), and use the decomposition to evaluate the integral /f(x)dx

3. Evaluate improper integral: from 0-infinity xe^-x dx(hint: use integration by parts)

4. Evaluate improper integral: from0-2 (dx/(sqrt(4-x^2))

5. The region R enclosed by the curves y-x/4 and y^2=x is rotated about the y-axis. Find the volume of the resulting solid.

Thank you very much for your help

2. Originally Posted by drmguy
1. Find the limit: lim x->0

(cos 3x)^1/x (Hint: use logarithmic method)

2. let f(x)=(5x^2 + 1)/((x-1)(x+1)^2) Find a,b,c such that f(x)=(a/(x-1))+(b/(x+1))+(c/(x+1)^2), and use the decomposition to evaluate the integral /f(x)dx

3. Evaluate improper integral: from 0-infinity xe^-x dx(hint: use integration by parts)

4. Evaluate improper integral: from0-2 (dx/(sqrt(4-x^2))

5. The region R enclosed by the curves y-x/4 and y^2=x is rotated about the y-axis. Find the volume of the resulting solid.

Thank you very much for your help
3. I don't like integration by parts, I prefer to use the "go in reverse twice" method.

$\frac{d}{dx}(xe^{-x}) = e^{-x} - xe^{-x}$

So $\int{e^{-x} - xe^{-x}\,dx} = xe^{-x} + C$

$\int{e^{-x}\,dx} - \int{xe^{-x}\,dx} = xe^{-x} + C$

$-e^{-x} - \int{xe^{-x}\,dx} = xe^{-x} + C$

$\int{xe^{-x}\,dx} = -xe^{-x} - e^{-x} + C$

Now to evaluate this integral from $0$ to $\infty$...

$\int_0^\infty{xe^{x}\,dx} = \lim_{\varepsilon \to \infty}(-\varepsilon e^{-\varepsilon} - e^{-\varepsilon} + C) - (-0e^{-0} - e^{-0} + C)$

Can you go from there?

3. Originally Posted by drmguy
1. Find the limit: lim x->0

(cos 3x)^1/x (Hint: use logarithmic method)

2. let f(x)=(5x^2 + 1)/((x-1)(x+1)^2) Find a,b,c such that f(x)=(a/(x-1))+(b/(x+1))+(c/(x+1)^2), and use the decomposition to evaluate the integral /f(x)dx

3. Evaluate improper integral: from 0-infinity xe^-x dx(hint: use integration by parts)

4. Evaluate improper integral: from0-2 (dx/(sqrt(4-x^2))

5. The region R enclosed by the curves y-x/4 and y^2=x is rotated about the y-axis. Find the volume of the resulting solid.

Thank you very much for your help
4. $\int_0^2{\frac{1}{\sqrt{4- x^2}}\,dx} = \left[\arcsin{\left(\frac{x}{2}\right)}\right]_0^2$

$= \lim_{\varepsilon \to 2}\left[\arcsin{\left(\frac{\varepsilon}{2}\right)}\right] - \arcsin{\left(\frac{0}{2}\right)}$

Can you go from here?

4. I am just worried that if i dont use integration by parts, as he requested, then i wont get full credit.