Question

\[\int\limits_{0}^{1}xe^{-x}~dx = \]a) 1-2e
b) -1
c) 1-2e^(-1)
d) 1
e) 2e-1

Answer 1

Could someone check my work pwease? :3

u = x v = -e^(-x)
du = dx dv = e^(-x) dx
uv - ∫v du
-xe^(-x) + ∫e^(-x) dx
-xe^(-x) - e^(-x)
-e^(-1) - e^(-1) - (-e^(0))
-2e^(-1) + 1

so C ?

Answer 2

\[I=\int\limits_{0}^{1}x~e ^{-x}dx=[ x \frac{ e ^{-x} }{ -1 }-\int\limits_{0}^{1} 1~\frac{ e ^{-x} }{ -1 }dx ]from~0\rightarrow1\]
\[=\left[ -\frac{ x }{ e^x }+\frac{ e ^{-x} }{ -1 } \right]~from~0\rightarrow1\]
\[=-\left[ \frac{ x+1 }{ e^x } \right]~from~0\rightarrow1\]
\[=-\left[ \frac{ 1+1 }{ e^1 }-\frac{ 0+1 }{ e^0 } \right]=-\frac{ 2 }{ e }+1=1-2 e ^{-1}\]
so you are correct.

Answer 3

Thank you! :)

Answer 4

yw