# Thread: Integration (finding the are bounded by a curve)

1. ## Integration (finding the are bounded by a curve)

To find the are bounded by the curve

F(x) = xe^(x^2) and the limits a=-2 and b=-1

So I integrate the function using substitution

$u=x$ and get stuck on how to go on

Help much appreciated!! thanks!!!

2. Originally Posted by Yehia
To find the are bounded by the curve

F(x) = xe^(x^2) and the limits a=-2 and b=-1

So I integrate the function using substitution

$u=x$ and get stuck on how to go on

Help much appreciated!! thanks!!!
I think you are on the wrong track. Look at the function:

$xe^{x^2}$

Now integrate it simply, don't use any U-Substitution.

$\int{xe^{x^2}}dx = \frac{1}{2}e^{x^2} + C$

I'll let you take it from there. If you still don't know how to do it, feel free to ask me.

3. Originally Posted by mathemagister
I think you are on the wrong track. Look at the function:

$xe^{x^2}$

Now integrate it simply, don't use any U-Substitution.

$\int{xe^{x^2}}dx = \frac{1}{2}e^{x^2} + C$

I'll let you take it from there. If you still don't know how to do it, feel free to ask me.

Aaahh no i see it now. Thanks a lot for your help. I see now why. NO substitution required.

4. Originally Posted by Yehia
Aaahh no i see it now. Thanks a lot for your help. I see now why. NO substitution required.
Anytime

5. Originally Posted by mathemagister
Anytime
Hi, sorry could you help me with just one more problem?
If i wish to find the Volume of a curve rotated round the x axis. i know it is the integral of pix^2, but in the funtion: $f(x)=x^(1/2)e^x$ with the limits 1 and 2, i am stuck

6. Originally Posted by Yehia
Hi, sorry could you help me with just one more problem?
If i wish to find the Volume of a curve rotated round the x axis. i know it is the integral of pix^2, but in the funtion: $f(x)=x^(1/2)e^x$ with the limits 1 and 2, i am stuck
The formula for the volume of any function rotated around $y=k$ is given by $V=\int_a^b \pi (R^2 - r^2) dx$ where a and b are the limits of integration, R (outer radius) is the distance from the axis of rotation to the further side of the elemental strip and r (inner radius) is the distance from the axis of rotation to the closer side of the elemental strip. In your case, since the axis of rotation is the x-axis, which is $y=0$, your r is simply 0. This will cancel out and your formula will become $V=\int_a^b \pi R^2 dx$.

$V=\int_1^2 \pi (f(x))^2 dx$

Hope I helped
Do you get it now?

7. Originally Posted by mathemagister
The formula for the volume of any function rotated around $y=k$ is given by $V=\int_a^b \pi (R^2 - r^2) dx$ where a and b are the limits of integration, R (outer radius) is the distance from the axis of rotation to the further side of the elemental strip and r (inner radius) is the distance from the axis of rotation to the closer side of the elemental strip. In your case, since the axis of rotation is the x-axis, which is $y=0$, your r is simply 0. This will cancel out and your formula will become $V=\int_a^b \pi R^2 dx$.

$V=\int_1^2 \pi (f(x))^2 dx$

Hope I helped
Do you get it now?
Yeah I get how to do it normally, it's just this specific one that i don't get.

the function is (x^0.5)(e^x) and the limits are 1 and 2 :S

8. Originally Posted by Yehia
Yeah I get how to do it normally, it's just this specific one that i don't get.

the function is (x^0.5)(e^x) and the limits are 1 and 2 :S
Oh you wrote the function weirdly before with LaTex and I wasn't sure what you meant.

$\int_1^2 \pi e^x \sqrt{x} dx$

Well, this is actually a non-elementary integral. This is impossible to compute normally. If you are more advanced in calculus you can do it, though. Do you know Dawson's function?

Well the answer to the indefinite integral is $e^x(\sqrt{x}-F(\sqrt{x}) + C$ where F is the Dawson Function. Yeah, a first (or even second) year calc student would not be responsible for being able to do this integral. Are you sure you weren't allowed to use a graphing calculator?