# Thread: Another Improper Integral

1. ## Another Improper Integral

I'm trying to show that $\int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=
\frac{32}{3}$

Since the integrand $\frac{1}{\sqrt[4]{x+2}}$ is continuous on $(-2,14]$ I can define

$\int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=\lim_{t->-2^+}\int_{t}^{14}\frac{dx}{\sqrt[4]{x+2}}$

$=\lim_{t->-2^+}\int_{t}^{14}(x+2)^{-\frac{1}{4}}dx$

$=\lim_{t->-2^+}-\frac{4}{3\sqrt[4]{(x+2)^4}}\displaystyle{]_{x=t}^{x=14}}$

$=-\frac{1}{4}+\lim_{t->-2^+}\frac{4}{3\sqrt[4]{(t+2)^3}}$

I'm stuck on the limit. If I didn't know any better, I would say that the integral diverges. I tried writing $\lim_{t->-2^+}\frac{4\sqrt[4]{(t+2)^3}}{3(t+2)^3}$ so I could use L-hopitals rule, but it doesn't seem to go anywhere since the derivative of the numerator always give me a function of the form $\frac{c}{f(t)}$ where $f(-2)=0$. So what's the trick here?

2. Originally Posted by adkinsjr
I'm trying to show that $\int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=
\frac{32}{3}$

Since the integrand $\frac{1}{\sqrt[4]{x+2}}$ is continuous on $(-2,14]$ I can define

$\int_{-2}^{14}\frac{dx}{\sqrt[4]{x+2}}=\lim_{t->-2^+}\int_{t}^{14}\frac{dx}{\sqrt[4]{x+2}}$

$=\lim_{t->-2^+}\int_{t}^{14}(x+2)^{-\frac{1}{4}}dx$

$=\lim_{t->-2^+}-\frac{4}{3\sqrt[4]{(x+2)^4}}\displaystyle{]_{x=t}^{x=14}}$

$=-\frac{1}{4}+\lim_{t->-2^+}\frac{4}{3\sqrt[4]{(t+2)^3}}$

I'm stuck on the limit. If I didn't know any better, I would say that the integral diverges. I tried writing $\lim_{t->-2^+}\frac{4\sqrt[4]{(t+2)^3}}{3(t+2)^3}$ so I could use L-hopitals rule, but it doesn't seem to go anywhere since the derivative of the numerator always give me a function of the form $\frac{c}{f(t)}$ where $f(-2)=0$. So what's the trick here?
$\lim_{t\to-2^+}\int_{t}^{14} (x+2)^{-1/4}\,dx=\lim_{t\to-2^+}\tfrac{4}{3}\left.\left[(x+2)^{3/4}\right]\right|_{t}^{14}$.

I'm sure you can take it from here...

3. oh ok, no wonder I was having such a hard time with this one. I don't know why I decided that the exponent was negative. Thanks for your help.