# help with basic integral problem?

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• October 2nd 2009, 06:09 PM
yoman360
help with basic integral problem?
$\int_{-5}^5 \frac{2}{x^3} dx$

the answer i got is 0.

the answer in the back of the book says: Does not exist

what did i do incorrectly?

This is what I did: $\int_{-5}^5 \frac{2}{x^3} dx$
= $\frac{-1}{x^2}|_{-5}^5$
= $\frac{-1}{(5)^2}-(\frac{-1}{(-5)^2})$
= $\frac{-1}{25}+\frac{1}{25}$
=0
• October 2nd 2009, 06:14 PM
mr fantastic
Quote:

Originally Posted by yoman360
$\int_{-5}^5 \frac{2}{x^3} dx$

the answer i got is 0.

the answer in the back of the book says: Does not exist

what did i do incorrectly?
Note* this post will be updated soon so i can show you my work

The update will be unnecessary since you probably neglected to recognise it as an improper integral (the integrand is undefined at x = 0 which lies in the interval of integration). The first step is therefore to write:

$\lim_{\alpha \rightarrow 0} \int_{-5}^{\alpha}\frac{2}{x^3} \, dx + \lim_{\beta \rightarrow 0} \int_{\beta}^{5}\frac{2}{x^3} \, dx$.

(A pre-emptive) by the way, statements like $\infty - \infty$ make no sense and certainly do not equal zero ....
• October 2nd 2009, 06:20 PM
yoman360
Quote:

Originally Posted by mr fantastic
The update will be unnecessary since you probably neglected to recognise it as an improper integral (the integrand is undefined at x = 0 which lies in the interval of integration). The first step is therefore to write:

$\lim_{\alpha \rightarrow 0} \int_{-5}^{\alpha}\frac{2}{x^3} \, dx + \lim_{\beta \rightarrow 0} \int_{\beta}^{5}\frac{2}{x^3} \, dx$.

(A pre-emptive) by the way, statements like $\infty - \infty$ make no sense and certainly do not equal zero ....

ah i see since f is not continuous on [a,b] I can't use the fundamental theorem of calculus:
$\int_a^b f(x)dx=F(b)-F(a)$

where F is any antiderivative of f, that is, a function such that F'=f

Thank you so much for pointing that out I just forgot to check if its continuous on [-5,5] since its discontinuous at x=0 the integral does not exist
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