Why does ∫(e^(-x))/(x^(1/2)) from 0 to infinity converge

I just got out of a calc II final, and have a few questions. We never get the final back, or see the grade, so this is basically for closure:

1. Why does ∫(e^(-x))/(x^(1/2)) from 0 to infinity converge?

2. what is ∫(x^3)/(sqrt(9-x^2))

3. What is the integral for the volume of the area, R, bounded by y= x+3 and y= x^2 -x rotated about the line x=4?

4. why does the infinite series ((n-10sqrt(n))/(n^2) converge?

5. what is the sum of the sequence from n=2 to 100 of (1/(n+1)) - (1/(n-1))

6. Solve the linear differential equation: (x^3)y' - (x^4)(y) = e^x

7. does the infinite series from n=1 to infinite ((1-(2/n)^(3n^2))(x^n) converge or diverge?



Any answers would be appreciated, I've been stressed about it all day, so a few answers would give me some closure.

Thanks

1) First, let z = x^(1/2) ==> x = z^2, dx = 2z dz.
==> ∫(z = 0 to ∞) 2 e^(-z^2) dz.
(Note that this removed one of the integral's singularities.)

Next, ∫(z = 0 to ∞) 2 e^(-z^2) dz
= ∫(z = 0 to 1) 2 e^(-z^2) dz + ∫(z = 1 to ∞) 2 e^(-z^2) dz

So, we only need to see if ∫(z = 1 to ∞) 2 e^(-z^2) dz converges.

This is done via comparison test:
2 e^(-z^2) < 2 e^(-z) for all z > 1.

Since ∫(z = 1 to ∞) 2 e^(-z) dz = -2e^(-z) {for z = 1 to ∞} = 2 (convergent),
the integral in question also converges.
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2) Use trig substitution x = 3 sin t, dx = 3 cos t dt.
==> ∫ (27 sin^3(t)) * (3 cos t dt) / (3 cos t)
= 27 ∫ (1 - cos^2(t)) sin t dt
= 27 (-cos t + cos^3(t)/3) + C
= (-27 cos t + 9 cos^3(t)) + C
= -27(sqrt(9 - x^2)/3) + 9 (sqrt(9 - x^2)/3)^3 + C
= -9 sqrt(9 - x^2) + (1/3) (9 - x^2)^(3/2) + C.
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3) Since x = 4 is parallel to the y-axis, I'll use cylindrical shells.

Note that x + 3 = x^2 - x
==> x^2 - 2x - 3 = 0
==> x = 3, -1.

On [-1, 3], we have x + 3 > x^2 - x.