Question

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

Answer 1

Answer:

$e^4+e^2$

Step-by-step 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.

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

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