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Find the area between y=(x-2)ex and the x-axis from x=2 to x=4.

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Answer:


e^4+e^2

Explanation:

It can be found by integral.

First let's find the intersection points.


(x-2)e^x = 0

Only intersection point is x = 2.

And it is asked to find the area in the interval (2, 4).


\int\limits^4_2 (x-2)e^x \,dx=?

We will use integration by parts.


x-2=u\\dx=du\\e^xdx=dv\\e^x=v


\int\limits^4_2 (x-2)e^x \,dx=uv-\int vdu=(x-2)e^x-\int e^xdx=(x-2)e^x-e^x=\\\\=(x-3)e^x|^4_2=(4-3)e^4-(2-3)e^2=e^4+e^2

User Vdudouyt
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