Do you only change the limits of integration when substituting? I get confused and don't know when to change the limits of integration.

2. Originally Posted by FalconPUNCH!
Do you only change the limits of integration when substituting? I get confused and don't know when to change the limits of integration.
You change the limit of an integral when you have a change of variables.

3. Originally Posted by polymerase
You change the limit of an integral when you have a change of variables.
Can you give me some examples?

4. Originally Posted by FalconPUNCH!
Can you give me some examples?
Sure.

Find $\int^4_2 \frac{2x+4}{x^2+4x+5}\,dx$

Now, in order to find this, you must first solve the intergrand. So one way to solve this question is to use the technique of substitution.

So I'm going to let $u=x^2+4x+5$. I have introduced a new variable $u$ in place of $x$. In the orginal function, the limit was x from 2 to 4 but now my intergral is in terms of $u$, which MAY OR MAY NOT go from 2 to 4. Thus to find out what it does go to, you plug in $x=2$ into the equation $u=x^2+4x+5$ to find out what $u$ goes to when $x$ goes to 2. And you do the same for the other limit, 4. In the end, you should find the new limits of u to be from 17 to 37.

Thus, $\int^4_2 \frac{2x+4}{x^2+4x+5}\,dx=\int^{37}_{17} \frac{du}{u}$

You then easily find the answer to be $\ln \left(\frac{37}{17}\right)$ since $\int^{37}_{17} \frac{du}{u}=\ln|u||^{37}_{17}$. You can also after solving the integral, re-substitute then solve again in terms of x from 2 to 4. Same answer!

Hope this helps!

5. Originally Posted by polymerase
Sure.

Find $\int^4_2 \frac{2x+4}{x^2+4x+5}\,dx$

Now, in order to find this, you must first solve the intergrand. So one way to solve this question is to use the technique of substitution.

So I'm going to let $u=x^2+4x+5$. I have introduced a new variable $u$ in place of $x$. In the orginal function, the limit was x from 2 to 4 but now my intergral is in terms of $u$, which MAY OR MAY NOT go from 2 to 4. Thus to find out what it does go to, you plug in $x=2$ into the equation $u=x^2+4x+5$ to find out what $u$ goes to when $x$ goes to 2. And you do the same for the other limit, 4. In the end, you should find the new limits of u to be from 17 to 37.

Thus, $\int^4_2 \frac{2x+4}{x^2+4x+5}\,dx=\int^{37}_{17} \frac{du}{u}$

You then easily find the answer to be $\ln \left(\frac{37}{17}\right)$ since $\int^{37}_{17} \frac{du}{u}=\ln|u||^{37}_{17}$. You can also after solving the integral, re-substitute then solve again in terms of x from 2 to 4. Same answer!

Hope this helps!
Oh so I have been doing them right I wasn't sure but now I get it. Thank You