What is the integral from-3 to x^2 of (t-5)(t+1) ?

The integral from-3 to x^2 of (t-5)(t+1) is -2x^4-5x^2+(x^6)/3+12

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integral from-3 to x^2 of (t-5)(t+1)

\int_{- 3}^{x^{2}} (t - 5) (t + 1) d t

- 2 x^{4} - 5 x^{2} + \frac{x ^{6}}{3} + 12

Solution steps

\int_{- 3}^{x^{2}} (t - 5) (t + 1) d t

Expand

(t - 5) (t + 1) : t^{2} - 4 t - 5

= \int_{- 3}^{x^{2}} t^{2} - 4 t - 5 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_{- 3}^{x^{2}} t^{2} d t - \int_{- 3}^{x^{2}} 4 t d t - \int_{- 3}^{x^{2}} 5 d t

\int_{- 3}^{x^{2}} t^{2} d t = \frac{x ^{6}}{3} + 9

\int_{- 3}^{x^{2}} 4 t d t = 2 (x^{4} - 9)

\int_{- 3}^{x^{2}} 5 d t = 5 x^{2} + 15

= \frac{x ^{6}}{3} + 9 - 2 (x^{4} - 9) - (5 x^{2} + 15)

Simplify

\frac{x ^{6}}{3} + 9 - 2 (x^{4} - 9) - (5 x^{2} + 15) : - 2 x^{4} - 5 x^{2} + \frac{x ^{6}}{3} + 12

= - 2 x^{4} - 5 x^{2} + \frac{x ^{6}}{3} + 12

What is the integral from-3 to x^2 of (t-5)(t+1) ?
The integral from-3 to x^2 of (t-5)(t+1) is -2x^4-5x^2+(x^6)/3+12