1. Integration by substitution

Problem: evaluate the integral from 4 to 9 of 1/x1/2+4 dx

2. Re: Integration by substitution

Originally Posted by Kvandesterren
Problem: evaluate the integral from 4 to 9 of 1/x1/2+4 dx
$\displaystyle \int_4^9 \frac{1}{\sqrt{x}} + 4 \, dx$

or

$\displaystyle \int_4^9 \frac{1}{\sqrt{x} + 4} \, dx$

3. Re: Integration by substitution

The second. Sorry, I haven't learned how to type math yet. How did you do that, btw?

4. Re: Integration by substitution

Originally Posted by Kvandesterren
The second. Sorry, I haven't learned how to type math yet. How did you do that, btw?
it's called latex ... there's a forum section here that covers it.

$\displaystyle u = \sqrt{x} + 4$

$\displaystyle \sqrt{x} = u - 4$

$\displaystyle du = \frac{1}{2\sqrt{x}} \, dx$

$\displaystyle 2\sqrt{x} \, du = dx$

$\displaystyle 2(u-4) \, du = dx$

substitute ...

$\displaystyle \int_6^7 \frac{2(u-4)}{u} \, du$

$\displaystyle \int_6^7 2 - \frac{8}{u} \, du$

5. Re: Integration by substitution

Originally Posted by Kvandesterren
Problem: evaluate the integral from 4 to 9 of 1/x1/2+4 dx
A good strategy for integration using a u substitution is to see if it's possible to write your function as a product of some "inner" function and this inner function's derivative.

Here we have \displaystyle \displaystyle \begin{align*} \int_4^9{\frac{1}{\sqrt{x} + 4}\,dx} \end{align*}. Since we have \displaystyle \displaystyle \begin{align*} \sqrt{x} \end{align*} inside our function, we will probably need to have this in our "inner" function that we make u. Which means we will need its derivative, \displaystyle \displaystyle \begin{align*} \frac{1}{2\sqrt{x}} \end{align*} as a factor in our integrand. So by manipulating the integrand slightly...

\displaystyle \displaystyle \begin{align*} \int_4^9{\frac{1}{\sqrt{x} + 4}\,dx} &= \int_4^9{\frac{2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x} + 4\right)}\,dx} \\ &= 2\int_4^9{\left(\frac{\sqrt{x}}{\sqrt{x} + 4}\right)\frac{1}{2\sqrt{x}}\,dx} \end{align*}

and now it's relatively easy to see that a substitution of \displaystyle \displaystyle \begin{align*} u = \sqrt{x} + 4 \implies du = \frac{1}{2\sqrt{x}}\,dx \end{align*} will work nicely. Note that when \displaystyle \displaystyle \begin{align*} x = 4, u = 6 \end{align*} and when \displaystyle \displaystyle \begin{align*} x = 9, u = 7 \end{align*}, and also that \displaystyle \displaystyle \begin{align*} \sqrt{x} = u - 4 \end{align*}, then the integral becomes

\displaystyle \displaystyle \begin{align*} 2\int_4^9{\left(\frac{\sqrt{x}}{\sqrt{x} + 4}\right)\frac{1}{2\sqrt{x}}\,dx} &= 2\int_6^7{\frac{u - 4}{u}\,du} \\ &= 2\int_6^7{1 - \frac{4}{u}\,du} \\ &= 2\left[u - 4\ln{|u|}\right]_6^7 \\ &= 2\left[\left(7 - 4\ln{7}\right) - \left(6 - 4\ln{6}\right)\right] \\ &= 2\left(1 + 4\ln{6} - 4\ln{7}\right) \\ &= 2 + 8\ln{6} - 8\ln{7}\end{align*}