Answer:
[tex](\bar x,\bar y)=(\frac{e^2+1}{e^2-1},\frac{e^2+1}{4} )[/tex]
Step-by-step explanation:
[tex]y=e^x,\:y=0,\:x=0,\:x=2[/tex] are given.
Firstly, we will find the area.
[tex]A = \int\limits^2_0e^x\:dx=e^2-1[/tex]
Secondly, we will find the x-coordinate of the centroid.
[tex]\bar x=\frac{1}{A} \int\limits^b_a x f(x)\:dx=\frac{1}{e^2-1}\int\limits^2_0 xe^x\:dx =\frac{1}{e^2-1}(xe^x-e^x)|^2_0=\frac{e^2+1}{e^2-1}[/tex]
Finally, we will find the y-coordinate of the centroid.
[tex]\bar y=\frac{1}{A} \int\limits^b_a \frac{1}{2} [f(x)]^2\:dx=\frac{1}{e^2-1}\frac{1}{2} \int\limits^2_0 e^{2x}\:dx =\frac{1}{2(e^2-1)}(\frac{e^{2x}}{2} )|^2_0=\frac{e^4-1}{4(e^2-1)}=\frac{e^2+1}{4}[/tex]
So the coordinates of centroid is:
[tex](\bar x,\bar y)=(\frac{e^2+1}{e^2-1},\frac{e^2+1}{4} )[/tex]
The sketch is given in the attachment.
