Show that $\int_{0}^{1} \frac{1}{x^2 + \sqrt{x}} dx$ 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 :

$\int_{a}^{\infty} \frac{1}{x^p} dx$ where $a>1$; If p $<=$ 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.

  • $\begingroup$ This may not be how you want to approach this problem, but you can compute that directly. You end up with a few logs and an arctangent - it's a bit of work but not terribly difficult. $\endgroup$ – DMcMor Jan 15 at 16:47
  • $\begingroup$ @DMcMor But at the lower limit of 0 the functions do not exist therefore they shouldn't be integrable according to the theory I have been taught. $\endgroup$ – Shaurya Goyal Jan 15 at 16:50
  • 1
    $\begingroup$ You can still find the antiderivative of $\frac{1}{x^2 + \sqrt{x}}$, say $F(x) = \int \frac{1}{x^2 + \sqrt{x}}\,dx$ and compute $F(1) - \lim_{x\to 0^{+}}F(x)$ to check convergence. $\endgroup$ – DMcMor Jan 15 at 17:05

You use the wrong test integral. Instead, you should use $$ \int_0^1\frac{1}{x^p}\,dx $$ for the cases $0<p<1$ and $p>1$.


You want to analyze the integral $\displaystyle\int_0^1\frac{1}{x^2+\sqrt{x}}dx$, which is fundamental different from integrals of the form $\displaystyle \int_1^\infty f(x)\, dx$. These are two different types of improper integrals.

In general, if $0\le f(x)\le g(x)$ and $\int_0^1g(x)dx$ is convergent, then $\int_0^1f(x)dx$ is also convergent.

In your example, $\displaystyle g(x)=\frac{1}{x^{1/2}}$.

  • $\begingroup$ I haven't been taught this. Could also add the cases as to when it is convergent and when it is diverging ? Im guessing at p=1 it diverges as log zero becomes infinite. But you havent included 1 in the range of p $\endgroup$ – Shaurya Goyal Jan 15 at 16:51
  • $\begingroup$ @ShauryaGoyal: Yes, you have parallel cases. The case when $p=1$ is also divergent. $\endgroup$ – user9464 Jan 15 at 16:57

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.



$$\frac{1}{x^2+\sqrt x}\leq \frac{1}{\sqrt x}.$$

  • $\begingroup$ I did try this only to realize that at the lower limit of 0 both of these functions do not exist. Thus they're not proper and not integrable. $\endgroup$ – Shaurya Goyal Jan 15 at 16:49
  • $\begingroup$ @ShauryaGoyal: I don't understand your comment. My hint allows you to prove that your integral converges. $\endgroup$ – Surb Jan 15 at 20:25

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