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-Find the area of the triangular region in the first quadrant that is bounded above by the curve y=e^(2x), below by y=e^x, and on the right by the line x=ln(3).

Thanks for any help

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-Find the area of the triangular region in the first quadrant that is bounded above by the curve y=e^(2x), below by y=e^x, and on the right by the line x=ln(3).

Thanks for any help

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Try plotting the curves, and check definite integrals.

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Science Advisor

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Science Advisor

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have you tried estimating it? i.e. can you show the area is less than 1,000?

or 10?

or 10?

Science Advisor

Homework Helper

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The graph of y= e

The area is [tex]\int_0^{ln 3}(e^{3x}- e^x)dx[/tex].

Can you do that integral?

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one point is (0,1)...trivial solution

put x=ln3 in e^x and get y=3........(ln3,3)

put x=ln3 in e^2x and get y=9........(ln3,9)

now, for the next part you have to draw the graph.

integrate e^x limits 0 to 3 to get e^3 - 1

integrate e^2x limits 0 to 9 to get (e^9 - 1)/2

The area required is [(e^9 - 1)/2 - (e^3 - 1)]

note: area under graph of a function between a and b is its definite integral with a and b as lower and upper limits.

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