help! integrating exponential (1 Viewer)

simDS · Jun 24, 2009

how do you find the integral of xe^x^2 bounderies x=1-2

thanks

kurt.physics · Jun 24, 2009

simDS said:
how do you find the integral of xe^x^2 bounderies x=1-2

thanks

Let $f(x) = x^2$ , note that $f'(x) = 2x$

Now look back at the integral

$\int_{1}^{2} xe^{x^2} \cdot dx$

Now one integral you have to remember is

$\int f'(x) e^{f(x)} dx = e^{f(x)} + C$

But comparing this to our integral, we have x but we need 2x, so in order to solve the integral, consider the following:

$\int_{1}^{2} \frac{1}{2} \times 2x e^{x^2} \cdot dx$

Now as $\frac{1}{2} \times 2x = x$

then the two integral are exactly the same except we can work on the one we found.

$\int_{1}^{2} \frac{1}{2} \times 2x e^{x^2} \cdot dx$

$= \frac{1}{2} \times \int_{1}^{2}2x e^{x^2} \cdot dx$

Now we know that:

$\int 2x e^{x^2} \cdot dx = e^{x^2}$

(because of the integral "formula" i mentioned)

so the problem becomes

$\frac{1}{2} \times \left[e^{x^2}\right]_{1}^{2}$

$= \frac{1}{2} \times \left(e^{4} - e\right)$

which is the answer, you could find an approximation but it isnt neccesary.

simDS · Jun 24, 2009

thank you so much!

Aquawhite · Jun 24, 2009

You're so quick to answer these kurt. You beat me once again >.<

kurt.physics · Jun 24, 2009

Aquawhite said:
You're so quick to answer these kurt. You beat me once again >.<

And yet you were still a worthy opponent .... good luck for battle to answer the next question

Aquawhite · Jun 24, 2009

I can really see you going insanely well in the HSC... You seem to a very intellectual and bright young child.

help! integrating exponential (1 Viewer)

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