Area between two curves problem

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Homework Help Overview

The problem involves finding the area between the curves defined by y = e^x, y = 2, and the y-axis. Participants are exploring how to set up the integral correctly and interpret the bounds related to the y-axis.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the integral, questioning how to incorporate the y-axis into their calculations. There are attempts to clarify the intersection points of the curves and the implications for the bounds of integration.

Discussion Status

The discussion is ongoing, with participants providing insights into the setup of the integral and the need to determine bounds. Some guidance has been offered regarding the relationship between the curves and the y-axis, but no consensus or final solution has been reached.

Contextual Notes

Participants are considering the implications of the y-axis as a boundary and are attempting to identify the intersection points of the curves to establish the limits for integration.

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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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