# What is the surface area of the solid created by revolving #f(x) = e^(2x) , x in [2,7]# around the x axis?

See the answer below:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the surface area of the solid created by revolving the function f(x) = e^(2x) on the interval [2, 7] around the x-axis, you can use the formula for the surface area of revolution:

[S = \int_{a}^{b} 2\pi f(x) \sqrt{1 + [f'(x)]^2} dx]

Where:

- (a) and (b) are the bounds of integration.
- (f(x)) is the function being revolved.
- (f'(x)) is the derivative of the function with respect to (x).

For the function (f(x) = e^(2x)):

- (f'(x) = 2e^(2x)).

Plugging in these values and integrating from (x = 2) to (x = 7):

[S = \int_{2}^{7} 2\pi e^(2x) \sqrt{1 + [2e^(2x)]^2} dx]

You would then solve this integral to find the surface area.

By signing up, you agree to our Terms of Service and Privacy Policy

- How do you find the volume of the solid generated by revolving the region bounded by the curves #y^2 = x*(4-x)^2# from x=0 to x=4 rotated about the x-axis?
- How long can exponential growth last?
- How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=8-x^2# #y=x^2# #x=0# rotated about the y axis?
- How do you find the average value of #f(x)=x/sqrt(x^2+1), 0<=x<=4#?
- How do you find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis #y=e^(-x)#, y=0, x=0, x=1?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7