Answer:
[tex]e^4+e^2[/tex]
Step-by-step explanation:
It can be found by integral.
First let's find the intersection points.
[tex](x-2)e^x = 0[/tex]
Only intersection point is x = 2.
And it is asked to find the area in the interval (2, 4).
[tex]\int\limits^4_2 (x-2)e^x \,dx=?[/tex]
We will use integration by parts.
[tex]x-2=u\\dx=du\\e^xdx=dv\\e^x=v[/tex]
[tex]\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[/tex]
