157k views
0 votes
Find the area between y=(x-2)ex and the x-axis from x=2 to x=4.

1 Answer

0 votes

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
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories