Improper Integrals: Infinite Limits

[Silvanoshei]

Some of these setups are very confusing to me. One in question is...

\(\displaystyle ∫_{6}^{\infty}\frac{xdx}{\sqrt{3+x^{2}}}\)

So... we're looking for a substitution that will allow a standard integral form right? Could you explain the \(\displaystyle \frac{1}{n+1}u^{n+1}\) for me, and make me understand why n= -1/2? Thanks.

 

[tkhunny]

Well, let's see...

\(\displaystyle \int\limits_{6}^{\infty}\dfrac{x\;dx}{\sqrt{3+x^{2}}}\)

\(\displaystyle u = 3+x^{2}\implies du = 2x\;dx\)

\(\displaystyle \dfrac{1}{2}\cdot\int\limits_{6}^{\infty}\dfrac{2x\;dx}{\sqrt{3+x^{2}}} = \dfrac{1}{2}\cdot\int\limits_{39}^{\infty}\dfrac{du}{\sqrt{u}} = \dfrac{1}{2}\cdot\int\limits_{39}^{\infty}u^{-1/2} du\)

Does it look like a -1/2, yet?

 

[pka]

Some of these setups are very confusing to me. One in question is...

\(\displaystyle ∫_{6}^{\infty}\frac{xdx}{\sqrt{3+x^{2}}}\)

So... we're looking for a substitution that will allow a standard integral form right? Could you explain the \(\displaystyle \frac{1}{n+1}u^{n+1}\) for me, and make me understand why n= -1/2? Thanks.

Try \(\displaystyle u=1+x^2 \text{ and }du=2 x dx\)

 

[DrPhil]

Some of these setups are very confusing to me. One in question is...

\(\displaystyle ∫_{6}^{\infty}\frac{xdx}{\sqrt{3+x^{2}}}\)

So... we're looking for a substitution that will allow a standard integral form right? Could you explain the \(\displaystyle \frac{1}{n+1}u^{n+1}\) for me, and make me understand why n= -1/2? Thanks.

The negative (1/2) power means reciprocal of a square root.

Let \(\displaystyle u = 3 + x^2\),......\(\displaystyle du = 2x\ dx\)

Then \(\displaystyle \dfrac{1}{\sqrt{3 + x^2}} = u^{-1/2}\), and \(\displaystyle x\ dx = du/2\)

ok? The other part you have to remember about the substitution is that the limits of the integral have to be changed from values of x to values of u.

\(\displaystyle x = 6 \implies u = 39,\;\;\;\;x = \infty \implies u = \infty\)

You will not be able to evaluate at the infinite limit.

 

[Silvanoshei]

Well, let's see...

\(\displaystyle \int\limits_{6}^{\infty}\dfrac{x\;dx}{\sqrt{3+x^{2}}}\)

\(\displaystyle u = 3+x^{2}\implies du = 2x\;dx\)

\(\displaystyle \dfrac{1}{2}\cdot\int\limits_{6}^{\infty}\dfrac{2x\;dx}{\sqrt{3+x^{2}}} = \dfrac{1}{2}\cdot\int\limits_{39}^{\infty}\dfrac{du}{\sqrt{u}} = \dfrac{1}{2}\cdot\int\limits_{39}^{\infty}u^{-1/2} du\)

Does it look like a -1/2, yet?

Why did the limit change to 39?

 

[DrPhil]

Why did the limit change to 39?

Remember when doing a substitution
that the limits of the integral have to be changed from values of x to values of u.

\(\displaystyle x=6 \implies u=39,\;\;\;\; x=\infty \implies u=\infty\)

 

[Silvanoshei]

Remember when doing a substitution
that the limits of the integral have to be changed from values of x to values of u.

\(\displaystyle x=6 \implies u=39,\;\;\;\; x=\infty \implies u=\infty\)

ohhhhh, you use u and plug 6 into x to get 39. Thanks!