Question

Find the exact coordinates of the centroid.

Answer 1

\[y=e^x, y=0, x=0, x=5\]

Answer 2

\[A=\int\limits_{0}^{5}e^xdx=[e^x]_{0}^{5}=e^5-1\]\[x = \frac{ 1 }{ A }\int\limits_{0}^{5}xe^xdx=\frac{ 1 }{ e^5-1 }[xe^x-e^x]_{0}^{5}\]\[\frac{ 1 }{ e^5-1 }[5e^5-e^5)-(0-1)] = \frac{ 1 }{ e^5-1 }[5e^5-e^5+1]\]\[y =\frac{ 1 }{ A }\int\limits_{0}^{5}\frac{ 1 }{ 2 }(e^x)^2dx=\frac{ 1 }{ e^5-1 }*\frac{ 1 }{ 4 }[e^{2x}]_{0}^{5}=\frac{ 1 }{ 4(e^5-1) }(e^{10}-1)=\frac{ e^5-1 }{ 4 }\]\[(x, y) = \frac{ 4e^5+1 }{ e^5-1 }, \frac{ e^5-1 }{ 4 })\]

The answer is wrong, what do?

Answer 3

I think that Area \(A\) is given by the subsequent formula:
\[\huge A = \int_0^5 {dx\int_0^{{e^x}} {dy} } \]

Answer 4

and:
\[\Large {x_G} = \frac{1}{A}\int_0^5 {xdx\int_0^{{e^x}} {dy} } \]

Answer 5

So, I should do these in terms of y -- not x?

Answer 6

no, no, I think your steps are right, since I got this:
\[\Large {x_G} = \frac{1}{A}\int_0^5 {xdx\int_0^{{e^x}} {dy} } = \frac{{4{e^5} + 1}}{{{e^5} - 1}}\]

Answer 7

here is the formula for \(y_G\):
\[\Large {y_G} = \frac{1}{A}\int_0^5 {dx\int_0^{{e^x}} {ydy} } \]
please try to develop such quantity

Answer 8

is that a double integral?

Answer 9

yes! Of course!

Answer 10

I haven't learn double integrals yet

Answer 11

I got this:
\[\Large {y_G} = \frac{1}{A}\int_0^5 {dx\int_0^{{e^x}} {ydy} } = \frac{{{e^{10}} - 1}}{{4\left( {{e^5} - 1} \right)}}\]
is it right?

Answer 12

Okay, I found my mistake. Turns out I didn't factor the "y" coordinate properly. \[\frac{ 1 }{ 4(e^5-1) }(e^{10}-1)=\frac{ 1 }{ 4(e^5-1) }(e^5-1)^2=\frac{ 1 }{ 4(e^5-1) }(e^5+1)(e^5-1)\]\[\frac{ e^5+1 }{ 4 }\]

Answer 13

Thank you for the help :)

Answer 14

:)