Area between two curves problem

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SUMMARY

The discussion focuses on calculating the area between the curves defined by the equations y = e^x, y = 2, and the y-axis. The correct approach involves setting up the integral as ∫ (e^x - 2) dy, with the bounds determined by the intersection of the curves and the y-axis. The lower bound is x = 0, while the upper bound is found by solving for the intersection of y = e^x and y = 2, which occurs at x = ln(2). The final area calculation yields 4ln(4) - 4.

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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?
 
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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?
 

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