I am an Exernal student (no in-person tuition or lectures), and I am practicing problems from my Calculus Study Guide, but the answers are not given. I'd appreciate if you could check my answers.

1. Determine whether the following integrals converge:

1.1 $\displaystyle \int^1_0\frac{cos x}{\sqrt{x}}dx$

Type 2 improper integral - problem spot: x=0

Since $\displaystyle \mid\cos x\mid\leq1$, compare the original integral to $\displaystyle \int^1_0\frac{1}{\sqrt{x}}dx$ which we know converges near 0 as the power of the denominator is less than 1.

Since $\displaystyle \int^1_0\frac{cos x}{\sqrt{x}}dx \leq \int^1_0\frac{1}{\sqrt{x}}dx$, which converges, the original integral also converges.

1.2 $\displaystyle \int^1_0\frac{\exp^{-x}}{\sqrt{x}}dx$

Type 2 improper integral - problem spot: x=0.

Choose a function $\displaystyle g(x)=\frac{1}{\sqrt {x}}$ which is similar to the original integrand $\displaystyle f(x)=\frac{\exp^{-x}}{\sqrt{x}}$ since

$\displaystyle lim_{x\rightarrow0}\frac{f(x)}{g(x)}=\exp^{(-x)}=1$

$\displaystyle \int^1_0\frac{1}{\sqrt{x}}dx$ converges, and so does the original integral.

1.3 $\displaystyle \int^{\infty}_1\frac{\exp^{-x}}{\sqrt{x\ln(x+1)}}dx$

Type 1 improper integral - problem spot: infinity.

Chose a comparision function $\displaystyle g(x)=\frac{\exp(-x)}{\sqrtx}}$

$\displaystyle lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}=\frac{1} {\sqrt{\ln{x+1}}}=0$ as the denominator goes to infinity

Since $\displaystyle \int^{\infty}_1\frac{\exp^{-x}}{\sqrt{x}}dx$ converges, so does the original integral, by one-way limit test.

1.4 $\displaystyle \int^1_0\frac{\exp^{-x}}{\sqrt{x\ln(x+1)}}dx$

Type 2 improper integral - problem spot: x=0.

Here I am a bit lost and would appreciate some hints. I know that the denominator is bounded (ln (x+1) ranges from 0 to ln2, and x ranges from 0 to 1. Does it help to chose a convergence test?

+I will post more in a follow up post.