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[SOLVED] Whitman 8.7.3 int with e

karush

Well-known member
Jan 31, 2012
2,724
$\tiny{s8.4.4.1\quad Whitman 8.7.3}$
Evaluate $\int(e^{t^2}+16)te^{t^2} dt$

ok I was going to distribute but not sure if a substitution would be better
the answer looks like it is done by observation


W|A returned

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skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
702
pretty easy to see the derivative of $(e^{t^2} + 16)$ is $2t e^{t^2}$ ... all one needs is the constant $2$ with $\dfrac{1}{2}$ outside the integral to set it up

$\displaystyle \dfrac{1}{2} \int (e^{t^2}+16) \cdot 2te^{t^2} \, dt = \dfrac{1}{2} \int u \, du$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
464
To integrate $\int (e^{t^2}+ 16)te^{t^2}dt$. It should be obvious that the "$t^2$ in $e^{t^2}$ is the "problem" since $e^{t^2}$ alone does not have an elementary anti-derivative. So cross your fingers and let "$u= t^2$". Then $du= 2t dt$. Alright! There is a "t" in the integrand!

We can the integral as $\int (e^{t^2}+ 16)e^{t^2}(tdt)$ and since, with the substitution $u= t^2$, $du= 2tdt$, so that $tdt= \frac{1}{2}du$ the integral becomes $\int (e^u+ 16)e^u (\frac{1}{2}du)= \frac{1}{2}\int (e^u+ 16)e^u du$. And, looking at that and remembering that at the derivative of $e^u$ is just $e^u$ again, make another substitution, $v= e^u$ o so that $dv= e^u du$ and the integral becomes $\frac{1}{2}\int (v+ 16)dv$.

(If I were very clever I would have thought of the substitution $v= e^{t^2}$ immediately but I have to do things step by step.)

That's easy to integrate! $\frac{1}{2}\int (v+ 16) dv=\frac{1}{2}(\frac{1}{2}v^2+ 16v)+ C= \frac{1}{4}v^2+ 8v+ C$ where "C" is the "constant of integration".

But now we have to go back to the original variable, t. Since $v= e^u$, $\frac{1}{4}v^2+ 8v+ C= \frac{1}{4}(e^u)^2+ 8e^u+ C= \frac{1}{4}e^{2u}+ 8e^u+ C$. And since $u= t^2$, $\frac{1}{4}e^{2u}+ 8e^u+ C= \frac{1}{4}e^{2t^2}+ 8e^{t^2}+ C$.