What is the integral from-2 to x^2 of (t+4)(t-2) ?

The integral from-2 to x^2 of (t+4)(t-2) is x^4-8x^2+(x^6+8)/3-20

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integral from-2 to x^2 of (t+4)(t-2)

\int_{- 2}^{x^{2}} (t + 4) (t - 2) d t

x^{4} - 8 x^{2} + \frac{x ^{6} + 8}{3} - 20

Solution steps

\int_{- 2}^{x^{2}} (t + 4) (t - 2) d t

Expand

(t + 4) (t - 2) : t^{2} + 2 t - 8

= \int_{- 2}^{x^{2}} t^{2} + 2 t - 8 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_{- 2}^{x^{2}} t^{2} d t + \int_{- 2}^{x^{2}} 2 t d t - \int_{- 2}^{x^{2}} 8 d t

\int_{- 2}^{x^{2}} t^{2} d t = \frac{x ^{6} + 8}{3}

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

\int_{- 2}^{x^{2}} 8 d t = 8 x^{2} + 16

= \frac{x ^{6} + 8}{3} + x^{4} - 4 - (8 x^{2} + 16)

Simplify

= x^{4} - 8 x^{2} + \frac{x ^{6} + 8}{3} - 20

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What is the integral from-2 to x^2 of (t+4)(t-2) ?
The integral from-2 to x^2 of (t+4)(t-2) is x^4-8x^2+(x^6+8)/3-20