What is the integral from-1 to 2 of (2x-3)(x^2+7) ?

The integral from-1 to 2 of (2x-3)(x^2+7) is -87/2

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integral from-1 to 2 of (2x-3)(x^2+7)

\int_{- 1}^{2} (2 x - 3) (x^{2} + 7) d x

- \frac{87}{2}

Decimal

- 43.5

Solution steps

\int_{- 1}^{2} (2 x - 3) (x^{2} + 7) d x

Expand

(2 x - 3) (x^{2} + 7) : 2 x^{3} + 14 x - 3 x^{2} - 21

= \int_{- 1}^{2} 2 x^{3} + 14 x - 3 x^{2} - 21 d x

Apply the Sum Rule:

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

= \int_{- 1}^{2} 2 x^{3} d x + \int_{- 1}^{2} 14 x d x - \int_{- 1}^{2} 3 x^{2} d x - \int_{- 1}^{2} 21 d x

\int_{- 1}^{2} 2 x^{3} d x = \frac{15}{2}

\int_{- 1}^{2} 14 x d x = 21

\int_{- 1}^{2} 3 x^{2} d x = 9

\int_{- 1}^{2} 21 d x = 63

= \frac{15}{2} + 21 - 9 - 63

Simplify

\frac{15}{2} + 21 - 9 - 63 : - \frac{87}{2}

= - \frac{87}{2}

What is the integral from-1 to 2 of (2x-3)(x^2+7) ?
The integral from-1 to 2 of (2x-3)(x^2+7) is -87/2