How do you integrate $f(x)=-2t^2+3t-6$ using the power rule?

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Answer 1 · 2017-01-13T13:24:20

$int (-2t^2+3t-6)dt =-2/3t^3+3/2t^2-6t+C$

First we use the linearity of the integral:

$int (-2t^2+3t-6)dt = -2int t^2dt +3int tdt -6int dt$

Then the power rule states that:

$d/(dt) t^n = nt^(n-1) <=> int t^ndt = t^(n+1)/(n+1) + C$

So:

$int t^2dt = t^3/3$

$int tdt = t^2/2$

$int dt = t$

and putting it together:

$int (-2t^2+3t-6)dt =-2/3t^3+3/2t^2-6t+C$

How do you integrate f(x)=-2t^2+3t-6f(x)=-2t^2+3t-6 using the power rule?