Show that $\displaystyle \int_{0}^{1} \frac{1}{x^2 + \sqrt{x}} dx$ converges.
My method : This is a type of improper integral where the function becomes unbounded at the lowe limit of integration and both the limits of integration are finite. I took the given function as $f(x)$ being integrated from $0$ to $1$ and another function $g(x) = \frac{1}{\sqrt{x}}$ by taking $\frac{1}{\sqrt{x}}$ common from $f(x)$ and observing that at $x=0$ the convergence and divergence of the function $f(x) $ was solely dependent by this $\frac{1}{\sqrt{x}}$
I was taught a limit test to find the convergence or the divergence. So it says that when the function is of the unbounded type with finite bounds to integrate :
Find $\lim_{x \to 0 } \frac{f(x)}{g(x)}$ which I get as 1 in this question. Since the limit exists and is not equal to zero, whatever is the behaviour of $g(x)$, that is the behaviour of $f(x) $ also. Therefore by using the test integral I find that $g(x)$ diverges therfore $f(x) $ should also diverges. But the answer says that it converges. How is this possible ?
Test integral used :
$\displaystyle \int_{a}^{\infty} \frac{1}{x^p} dx$ where $a>1$;
If p $\leq$ 1 then it diverges.
If p $>$ 1 than it converges.
I know that in the test integral $a>1$ is given but then my prof used it irrespective of it in other examples of the same type (that is in sums of unbounded type where the limits of integration were $0$ to $3$ and $0$ to $4\pi$ in the two examples he gave) . And I also do not understand the $\infty$ limit in the integration of test integral as in the given question I need to integrate only from $0$ to $1$
I know I have asked two questions but since they're related please don't close my answer.
Best Answer
The integral of $g(x)$ is $\displaystyle \int_0^1 \frac{1}{\sqrt{x}}\,dx = 2\sqrt{x}\Big\rvert_0^1 = 2$, doesn't diverge.