What is the integral from-4 to 4 of (x^2+x-6) ?

The integral from-4 to 4 of (x^2+x-6) is -16/3

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integral from-4 to 4 of (x^2+x-6)

\int_{- 4}^{4} (x^{2} + x - 6) d x

- \frac{16}{3}

Decimal

- 5.33333 \dots

Solution steps

\int_{- 4}^{4} x^{2} + x - 6 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_{- 4}^{4} x^{2} d x + \int_{- 4}^{4} x d x - \int_{- 4}^{4} 6 d x

\int_{- 4}^{4} x^{2} d x = \frac{128}{3}

\int_{- 4}^{4} x d x = 0

\int_{- 4}^{4} 6 d x = 48

= \frac{128}{3} + 0 - 48

Simplify

\frac{128}{3} + 0 - 48 : - \frac{16}{3}

= - \frac{16}{3}

\int_{4}^{9} x x - 3 d x

\int_{2}^{4} 16 - p^{2} d p

\int_{0}^{1} y^{2} \cdot 3 (1 - y)^{2} d y

\int_{0}^{π} x^{2} cos (3 x) d x

\int_{0}^{π} cos^{4} (θ) d θ

What is the integral from-4 to 4 of (x^2+x-6) ?
The integral from-4 to 4 of (x^2+x-6) is -16/3