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

The integral from-4 to 4 of (-x^2+16) is 256/3

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integral from-4 to 4 of (-x^2+16)

\int_{- 4}^{4} (- x^{2} + 16) d x

\frac{256}{3}

Decimal

85.33333 \dots

Solution steps

\int_{- 4}^{4} - x^{2} + 16 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} 16 d x

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

\int_{- 4}^{4} 16 d x = 128

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

Simplify

- \frac{128}{3} + 128 : \frac{256}{3}

= \frac{256}{3}

What is the integral from-4 to 4 of (-x^2+16) ?
The integral from-4 to 4 of (-x^2+16) is 256/3