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#