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This is from James Stewart's Multivariable Calculus book, page 651 question 33. Find the area enclosed by the x-axis and the curve $x=1+e^t$, $y=t-t^2$.

The answer given has the following steps that I do not understand. It's supposed to be integration by parts. $$\int_0^1(t-t^2)e^tdt$$ $$=\int_0^1te^tdt-\int_0^1t^2e^tdt=\int_0^1te^tdt-[t^2e^t]_0^1+2\int_0^1te^tdt$$

My confusion is on $[t^2e^t]_0^1$, which is supposed to be $[f(x)g(x)]_a^b$ of the integration by part formula. Isn't $f(x)=t-t^2$ and $g(x)=1+e^t$?

Please advice.

user95087
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1 Answers1

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You appear to be confused with just the standard integration by parts formula from calculus of a single variable. Below is what integration by parts tells us. $$\int _a^bu\,dv=uv\bigg\vert_a^b-\int_a^b v\,du$$ Here we have $$ \int_0^1\underbrace{\color{red}{t^2}}_u\,\underbrace{\color{blue}{e^t\,dt}}_{dv} $$Since $dv=e^t\,dt$ we have $v=\int e^t\,dt=e^t$ hence $uv=t^2e^t$ as given.

In other words, the $f(t),g(t)$ come from our integrand:$$\int_a^b f(t)g'(t)dt=\left[f(t)g(t)\right]_a^b-\int f'(t)g(t)\,dt\\\int_a^b t^2e^t\,dt=[t^2e^t]_0^1-\int 2te^t\,dt$$Here, you have $f(t)=t^2,g'(t)=e^t$ and therefore $f'(t)=2t,g(t)=e^t$, not the unrelated functions $t-t^2,1+e^t$

obataku
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