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

The integral from-3 to x^2 of (t-5)(t^2+1) is (x^4-9)/2+(x^8-81)/4-5x^2-5*(x^6)/3-60

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

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

\frac{x ^{4} - 9}{2} + \frac{x ^{8} - 81}{4} - 5 x^{2} - 5 \cdot \frac{x ^{6}}{3} - 60

Solution steps

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

Expand

(t - 5) (t^{2} + 1) : t^{3} + t - 5 t^{2} - 5

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

\int_{- 3}^{x^{2}} t^{3} d t = \frac{x ^{8} - 81}{4}

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

\int_{- 3}^{x^{2}} 5 t^{2} d t = 5 (\frac{x ^{6}}{3} + 9)

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

= \frac{x ^{8} - 81}{4} + \frac{x ^{4} - 9}{2} - 5 (\frac{x ^{6}}{3} + 9) - (5 x^{2} + 15)

Simplify

\frac{x ^{8} - 81}{4} + \frac{x ^{4} - 9}{2} - 5 (\frac{x ^{6}}{3} + 9) - (5 x^{2} + 15) : \frac{x ^{4} - 9}{2} + \frac{x ^{8} - 81}{4} - 5 x^{2} - 5 \cdot \frac{x ^{6}}{3} - 60

= \frac{x ^{4} - 9}{2} + \frac{x ^{8} - 81}{4} - 5 x^{2} - 5 \cdot \frac{x ^{6}}{3} - 60

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What is the integral from-3 to x^2 of (t-5)(t^2+1) ?
The integral from-3 to x^2 of (t-5)(t^2+1) is (x^4-9)/2+(x^8-81)/4-5x^2-5*(x^6)/3-60