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Evaluating Definite integrals

shamieh

Active member
Sep 13, 2013
539
Evaluate the following integrals.

a) $\int^1_0 x e^x dx$

So integrating by parts we get

$u = x $ $vu = e^x dx$
$du = dx$ $ v = e^x$

$uv - \int vdu = x e^x - \int^1_0 e^x dx$

\(\displaystyle xe^x - e^x |^1_0 = 1\)

b) \(\displaystyle \int^1_0 x^2 e^x \, dx\)

Integrating by parts we get

\(\displaystyle u = x^2 \) \(\displaystyle dv = e^x dx\)
\(\displaystyle du = 2xdx\) \(\displaystyle v = e^x\)

\(\displaystyle uv - \int vdu = x^2 e^x - \int^1_0 e^x 2x = e^1 - 2 \)
 
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Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
I would be more careful about notation but your results are correct. Congratulations! (Clapping)
 

shamieh

Active member
Sep 13, 2013
539
I would be more careful about notation but your results are correct. Congratulations! (Clapping)
Any idea why the definite integral calculator is somehow getting 2 - e? are they equivalent?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is a new approach that you might like

Define

\(\displaystyle f(x) = \int^1_0 e^{xt} \, dt \)

Then integrating with respect to $t$ we have

\(\displaystyle f(x) = \int^1_0 e^{xt} \, dt = \frac{e^{x}-1}{x}\)

Now differentiate both sides with respect ot $x$ we have

\(\displaystyle f'(x) = \int^1_0 t\, e^{xt} \, dt = \frac{xe^{x}-e^x+1}{x^2}\)

Putting $x=1$ we have

\(\displaystyle f'(1) = \int^1_0 t\, e^{t} \, dt = 1\)

Diff w.r.t to $x$ again we have

\(\displaystyle f''(x) = \int^1_0 t^2\, e^{xt} \, dt = \frac{x^2 \,e^{x}-2(xe^x-e^x +1)}{x^3}\)

\(\displaystyle f''(1) = \int^1_0 t^2\, e^{t} \, dt = e-2\)

Any idea why the definite integral calculator is somehow getting 2 - e? are they equivalent?
The answer $2-e<0$ is not possible since the integral must be positive.
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
No, [tex]\displaystyle \begin{align*} 2 - e = - \left( e - 2 \right) \end{align*}[/tex]. They are not equivalent.

e - 2 is definitely correct. You can tell because the function is always positive, so the area is always above the x axis and thus must be positive.
 
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shamieh

Active member
Sep 13, 2013
539
Nevermind I se what I did wrong