What is the integral of 10t^2+7t-9 ?

The integral of 10t^2+7t-9 is (10t^3)/3+(7t^2)/2-9t+C

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integral of 10t^2+7t-9

\int 10 t^{2} + 7 t - 9 d t

\frac{10 t ^{3}}{3} + \frac{7 t ^{2}}{2} - 9 t + C

Solution steps

\int 10 t^{2} + 7 t - 9 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 10 t^{2} d t + \int 7 t d t - \int 9 d t

\int 10 t^{2} d t = \frac{10 t ^{3}}{3}

\int 7 t d t = \frac{7 t ^{2}}{2}

\int 9 d t = 9 t

= \frac{10 t ^{3}}{3} + \frac{7 t ^{2}}{2} - 9 t

Add a constant to the solution

= \frac{10 t ^{3}}{3} + \frac{7 t ^{2}}{2} - 9 t + C

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What is the integral of 10t^2+7t-9 ?
The integral of 10t^2+7t-9 is (10t^3)/3+(7t^2)/2-9t+C