Indefinite Integral

**WartonMorton** · February 5th 2010, 06:36 AM

Find the Integral of (t^2 - a)(t^2 - b) * dt.

I tried combining to (t^4 -bt^2 - at^2 + ab) but this didn't give me much.

**General** · February 5th 2010, 06:58 AM

Originally Posted by WartonMorton

Find the Integral of (t^2 - a)(t^2 - b) * dt.

I tried combining to (t^4 -bt^2 - at^2 + ab) but this didn't give me much.

No, It gives.
$\int (t^4 -bt^2 - at^2 + ab) dt = \int t^4 dt - (a+b)\int t^2 dt + ab \int dt$ .

**dedust** · February 5th 2010, 07:02 AM

Originally Posted by WartonMorton

Find the Integral of (t^2 - a)(t^2 - b) * dt.

I tried combining to (t^4 -bt^2 - at^2 + ab) but this didn't give me much.

hi warton,.
now you only need to integrate it using the integral formula for polynomial,

$\int t^n ~dt = \frac{1}{n+1} t^{n+1}$ , as long as $n \not = -1$

**General** · February 5th 2010, 07:05 AM

Originally Posted by dedust

hi warton,.
now you only need to integrate it using the integral formula for polynomial,

$\int t^n ~dt = \frac{1}{n+1} t^{n+1}$ , as long as $n \not = -1$

Actually, Its name is "Power Rule".

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