Answer
$A=3-e$
Work Step by Step
The values of t when y=0 are found by
$y=0$
$t-t^{2}=0$
$t(1-t)=0$
$t=0$ and $t=1.$
For $t\in[0,1]$, y(t) is positive.
We find
$A=\displaystyle \int_{0}^{1}y(t)[x'(t)dt]=\int_{0}^{1}(t-t^{2})e^{t}dt$
$=\displaystyle \int_{0}^{1}te^{t}dt-\int_{0}^{1}t^{2}e^{t}dt$
From reference page 10, use the formula
$97.\displaystyle \quad \int u^{n}a^{au}du=\frac{1}{a}u^{n}e^{au}-\frac{n}{a}\int u^{n-1}e^{au}du$
$\displaystyle \int_{0}^{1}te^{t}dt=te^{t}|_{0}^{1}-e^{t}|_{0}^{1},$
$\displaystyle \int_{0}^{1}t^{2}e^{t}dt=t^{2}e^{t}|_{0}^{1}-2\int_{0}^{1}te^{t}dt=t^{2}e^{t}|_{0}^{1}-2(te^{t}|_{0}^{1}-e^{t}|_{0}^{1})$
$A= (e-0)-(e-1)-\{(e-0)-2[(e-0)-(e-1)]\}$
$= 1-(e-2)$
$=3-e$
