use the definition of an improper integral to evaluate the given integral.$\int_3^\infty\frac{1}{(x+2)}dx$ 
I try to solve the first question 
\begin{align}
\lim \limits_{t \to \infty} {\int_{3}^{t} \frac{1}{x+2}}dx
\end{align}
and my answer for it is 
$$\ln(\infty+2)-\ln(5)$$
However, now I'm stuck on question 2, on $a$, $b$ and don't know how to solve them 
 A: The volume of the solid from rotating the curve $f(x)=e^{-x}$ around the $x$ axis, is given by
$$\pi\int_0^\infty (e^{-x})^2\,dx=\pi\int_0^\infty e^{-2x}\,dx$$
The volume of the solid from rotating the curve $f(x)=e^{-x}$ around the $y$ axis, is given by
$$\pi\int_0^1 (-\log y)^2\,dy=\pi\int_0^1 \log^2 (y)\,dy$$
Now, can you determine whether these integrals converge or diverge?
A: As I stated in the comments, the answer to the first question is $\infty$, which is what you would have gotten had you worked out the answer a bit more. (Nitpicking)
The second question involves a concept called finding the volume by rotation.

Finding the volume of a solid revolution is a method of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is much like finding the area, but with an added component of rotating the area around a line of symmetry - usually the x or y axis.

For more information, follow the link.
I will work out the answer to part a) of question 2, it's your responsibility to read up on this if you don't know what's going on.
This is the graph.

Now, what they want to know is what is the volume of a shape generated by rotating $e^{-x}$ around the $x$-axis.
Here's a visual of a similar shape being created,

Here's a visual of the completed shape.

Note that the equation is asking for the volume of these shapes, so just imagine them as being filled in instead of hollow.
So, let's walk through the equations required to do this, I'm only giving a basic overview because it is midnight and I'm exhausted (This is what the link is for).
$$\pi \int_0^\infty \text{Outer_Radius}^2-\text{Inner_Radius}^2dx$$
$$=\pi \int_0^\infty e^{-2x}-0^2dx$$
$$=\pi \int_0^\infty e^{-2x}dx$$
The question is does this integral converge or diverge?
Part b) asks you to do the exact same thing, except in terms of $y$, not $x$.
So,
$$y=e^{-x}$$
$$ln(\frac{1}{y}) = x$$
Now we just put everything in terms of $y$ and rotate around the $y$-axis.
Here's a visual.

Note that the $y$-value for the equation doesn't extend beyond 1, so that's the upper limit of the integral.
$$\pi \int_0^1 \text{Outer_Radius}^2-\text{Inner_Radius}^2dy$$
$$=\pi \int_0^1 ln(\frac{1}{y})^2-0^2dy$$
$$=\pi \int_0^1 (-ln(y))^2dy$$
$$=\pi \int_0^1 ln^2(y)dy$$
Does this integral converge or diverge?
