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

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

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

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

\frac{256}{3}

Decimal

85.33333 \dots

Solution steps

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

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

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

= 128 - \frac{128}{3}

Simplify

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

= \frac{256}{3}

f (x) = x^{- 2}

\frac{d}{d x} (lo g_{10} (e^{x}))

\frac{d y}{d t} = - \frac{1}{2} y

x \to 0 + lim (\frac{x ^{2} - 1}{x})

\frac{2 x - 1}{x + 1}

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