# Area between two curves problem

• zeezey
In summary, the task is to find the area enclosed by the functions y = e^x, y = 2, and the y-axis. The approach is to use the formula for finding the area between a curve and the y-axis, which is given by ∫ x dy. The lower bound for x is determined by the y-axis, and the upper bound is the intersection of the given functions. In this case, the upper bound is found by setting y = e^x = 2 and solving for x. The definite integral is then evaluated to find the area, which is equal to 4ln(4) - 4.

## Homework Statement

Find the area of the indicated region.
Enclosed by y = e^x, y = 2, and the y-axis

## The Attempt at a Solution

So I the area should be integral of e^x - 2 which = e^x - 2x . I'm not sure what to do with the y-axis it mentions?

zeezey said:

## The Attempt at a Solution

So I the area should be integral of e^x - 2 which = e^x - 2x . I'm not sure what to do with the y-axis it mentions?

Firstly where does y=ex cut the y-axis?

You should then know the area between a curve and the y-axis is given by ∫ x dy. So if y=ex, how do you put x in terms of y?

umm I'm not sure you mean like dx = e^x dy ?

So you have e^x-2 as the integrand, as you mentioned. The mention of the y-axis is to put a lower bound on x (in this case, evaluating your integral from a-to-b by the FTC, the lower bound, a, is x=0)...then the upper bound would be the intersection of the functions. It's always helpful to draw out what you're asked for.

I'll leave it to you to find the upper bound and then evaluate your now definite integral.

So I got 4ln(4) - 4 . Is this correct?