# Thread: definate integrals

1. ## definate integrals

1. /int 2_1 (e^(1/x)/x^2)dx the answer is e-e^(1/2)
2. /int pi/6_0 tan 2x dx the answer is (1/2)ln 2
3. I have a question, for a definite that have the same number except one is positive and the other is negative (/int 2_-2), is there a special way or equation to find the answer? It goes something like /int 0_-2 (-anything)-/int 2_0 (anything) or something of the like. I hope this is enough infomation.

For 1.u=1/x, du=-1/x^2. Well, in this circumstance, I don't know the rest since if it needs to be multiplied or divided or not.

For 2. u=2x du=2 dx. tan u -> -ln|cos(u)|*(1/2) and i think this is where i may have gotten something wrong.

2. Well, I've looked these over, and I have a solution to number 1, but sorry, I don't know how to do number 2.

3. Originally Posted by driver327
1. /int 2_1 (e^(1/x)/x^2)dx the answer is e-e^(1/2)
2. /int pi/6_0 tan 2x dx the answer is (1/2)ln 2
3. I have a question, for a definite that have the same number except one is positive and the other is negative (/int 2_-2), is there a special way or equation to find the answer? It goes something like /int 0_-2 (-anything)-/int 2_0 (anything) or something of the like. I hope this is enough infomation.

For 1.u=1/x, du=-1/x^2. Well, in this circumstance, I don't know the rest since if it needs to be multiplied or divided or not.

For 2. u=2x du=2 dx. tan u -> -ln|cos(u)|*(1/2) and i think this is where i may have gotten something wrong.
So for 2 you have:

$
\int_0^{\pi/6} \tan(2x)~dx = \frac{1}{2}~\int_0^{\pi/3} \tan(u)~du = -\frac{1}{2} \left[ \ln(|cos(u)|) \right]_0^{\pi/3}
$
$
=-\frac{1}{2}\ln(1/2) = \frac{1}{2}\ln(2)
$

RonL

4. One question, under what circumstances does the top and bottom of the integral change? Other than that, i'm fine.

5. Originally Posted by driver327
One question, under what circumstances does the top and bottom of the integral change? Other than that, i'm fine.
When you change the variable in a definte integral the limits change according to the change of variable.

So if we have an integral from a to b of some function of x (say f(x)) with repect to x and we make the change of variable u=h(x), the new integral is with respect to u from u=h(a) to u=h(b).

RonL