3unit Integration help (1 Viewer)

imsleepyz · Mar 28, 2012

How do I do the second part of this question? Can someone help me please?

nightweaver066 · Mar 28, 2012

By product rule,
$\frac{d}{dx}(xe^x) = xe^x + e^x$

Rearranging this,
$\frac{d}{dx}(xe^x) - e^x = xe^x$

Integrating both sides with respect to x with limits x = 0 to 1

$\int ^1_0 xe^x dx = \int^1_0 \frac{d}{dx}(xe^x) - e^xdx$

$= [xe^x - e^x]^1_0$

$= e - e - (0 - 1)$

$= 1$

imsleepyz · Mar 28, 2012

Thanks nightweaver066 but I have a question... can you tell me how you can integrate d/dx (xe^x)? Does the integration sign and d/dx cancel out?

nightweaver066 · Mar 28, 2012

imsleepyz said:
Thanks nightweaver066 but I have a question... can you tell me how you can integrate d/dx (xe^x)? Does the integration sign and d/dx cancel out?

Remember that integration is the reverse of differentiation.

$\int\frac{d}{dx}[f(x)]dx = f(x) + c$

imsleepyz · Mar 28, 2012

nightweaver066 said:
Remember that integration is the reverse of differentiation.

$\int\frac{d}{dx}[f(x)]dx = f(x) + c$

Ohhh yeah I see, I see... whatever the derivative of f(x) is, when you integrate it, it is still f(x). Thank you very much nightweaver066 for your help and your help for my other post.

3unit Integration help (1 Viewer)

imsleepyz

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nightweaver066

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imsleepyz

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nightweaver066

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imsleepyz

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