finding areas with e

• Feb 9th 2010, 11:00 AM
mthomas1
finding areas with e
I know how to do area problems but not with e, so i would appreciate the help.

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y = e^(5 x)
y = e^(6 x)
x = 1

I'm not sure if i'm suppose to rotate it around x=1, which means i need to change it in terms of x. and then i need to find the points and integrate the square of the difference between the e's (but i don't know how to square those terms either)

Find the volume of the solid formed by rotating the region enclosed by

y = e^(3 x)+4
y = 0
x = 0
x = 0.1

for this problem i'm sure you just integrate pi * the square of y = e^(3 x)+4 from 0 to 0.1, but I don't know how to square y = e^(3 x)+4.

any help would be appreciated.
• Feb 9th 2010, 11:09 AM
mthomas1
ok i figured out the last one, but i dont understand how you integrate e^6x + 8e^3 +16.
• Feb 9th 2010, 11:13 AM
drumist
Quote:

Originally Posted by mthomas1
I know how to do area problems but not with e, so i would appreciate the help.

Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

y = e^(5 x)
y = e^(6 x)
x = 1

I'm not sure if i'm suppose to rotate it around x=1, which means i need to change it in terms of x. and then i need to find the points and integrate the square of the difference between the e's (but i don't know how to square those terms either)

You're asked to calculate an area, not a volume. There's no rotation. Just find the area between the curves. I'm guessing you can do this if you realize it's not a rotation problem.

An an aside though, this is the correct procedure for squaring something with an exponent: $\left(e^{6x}\right)^2 = e^{6x\cdot2} = e^{12x}$

Quote:

Find the volume of the solid formed by rotating the region enclosed by

y = e^(3 x)+4
y = 0
x = 0
x = 0.1

for this problem i'm sure you just integrate pi * the square of y = e^(3 x)+4 from 0 to 0.1, but I don't know how to square y = e^(3 x)+4.
That's correct if you are rotating about the x-axis (y=0), but you didn't say if that was the case. Anyway, this works similar to the first one:

$(e^{3x}+4)^2 ~=~ (e^{3x}+4)(e^{3x}+4)$

$=~ \left(e^{3x}\right)^2 + 2\cdot4e^{3x}+16 ~=~ e^{6x} + 8e^{3x}+16$
• Feb 9th 2010, 11:22 AM
drumist
In general, the integral of $e^{ax}$ where a is any constant is $\tfrac{1}{a}e^{ax}$. So, for example, the integral of $e^{6x}$ is $\tfrac{1}{6}e^{6x}$. If you think about what the derivative of $e^{ax}$ would look like, this makes sense.
• Feb 9th 2010, 12:51 PM
mthomas1
thank you for explaining the integration of the last. so now the last question is solved for me.

but now that i realize the first is an area problem and not an area problem, i'm even more confused. Looking at the sketch of the graphs it looks like it should be in the form of y, but then i don't understand what x=1 is in there for. i'm confused on how to set it up.