Solve for $p$ in $\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$ I did a question $\int_{0}^{1}\frac{1}{x^{\frac{1}{2}}}\,dx$, and evaluating this is divergent integral yes? Then as a general form $\int_{0}^{1} \frac{1}{x^p}\,dx$, $p \in \mathbb{R}$, what values of $p$ can give me $\int_{0}^{1} \frac{1}{x^p}\,dx = \frac{4}{3}$? This is a easy integral to calculate, make it $\int_{0}^{1}x^{-p}dx$ and calculate, etc. Then how do I get this to solve $p$? I am using the fundamental theorem of calculus and confused here.
 A: Actually,$$\int_0^1x^{-1/2}\,\mathrm dx=\left[2x^{1/2}\right]_{x=0}^{x=1}=2.$$
On the other hand,$$\int_0^1x^{-p}\,\mathrm dx=\left[\frac{x^{1-p}}{1-p}\right]_{x=0}^{x=1}=\frac1{1-p}.$$So, take $p=\frac14$.
A: $\int_0^1{\frac{1}{x^\frac{1}{2}}}dx$ is not divergent.
$\int_0^1{\frac{1}{x^\frac{1}{2}}}dx=\int_0^1 x^{-\frac{1}{2}}dx=2x^\frac{1}{2}|_0^1=2$
Similarly solving: $\int_0^1 \frac{1}{x^p}dx=\frac{4}{3}$
$\int_0^1 \frac{1}{x^p}dx=\frac{x^{p-1}}{1-p}|_0^1=
\frac{1}{1-p}=\frac{4}{3}$
So, $p=\frac{1}{4}$
A: First of all $$\int_{0}^{1}\frac{1}{x^\alpha}$$ is divergent if $\alpha>1$, so the first case cited by you (for $\alpha=\frac{1}{2}$) corresponds to convergence.
Now let's observe that:
$\displaystyle\int_{0}^{1}\frac{1}{x^\alpha}=\displaystyle\lim_{x\to 0^+} \int_{x}^{1}\frac{1}{x^\alpha}=\lim_{x\to 0^+} \frac{1}{1-\alpha}(1-x^{1-\alpha})=\frac{1}{1-\alpha}$ when $\alpha<1$.
So: $\displaystyle\frac{1}{1-\alpha}=\frac{4}{3}$ when $\displaystyle\alpha=\frac{1}{4}$
A: It's divergent for $p\leq 1$.   Otherwise by the fundamental theorem of calculus,  you get $\frac {x^{-p+1}} {-p+1}$.  Plug in your 1 and 0 and then set equal to 4/3 should let you solve for p.
A: Start with the antiderivative, assuming that $p\ne 1$ $$\int x^{-p}\,dx = \frac{1}{1-p}x^{1-p} + C.$$  Then it must be the case that $$\frac{1}{1-p}x^{1-p}\bigg|_{0}^{1} = \frac{1}{1-p}\left(1-\lim_{x\to 0}x^{1-p}\right).$$  Now, in order for $\lim_{x\to 0}x^{1-p}$ to exist we need $1-p>0,$ in which case the limit is zero.  So, we need two conditions:
\begin{align}
p&<1\\[5pt]
\frac{1}{1-p} &= \frac{4}{3}.
\end{align}
We can calculate $p$ from the second condition to get $p = 1/4$.  This also satisfies the condition that $p<1$.  To confirm that we have the correct result:
$$\int_{0}^{1}x^{-1/4}\,dx = \frac{4}{3}x^{3/4}\bigg|_{0}^{1} = \frac{4}{3} - 0 = \frac{4}{3}.$$
