What is the integral from 0 to 4 of (16-x^2) ?

The integral from 0 to 4 of (16-x^2) is 128/3

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integral from 0 to 4 of (16-x^2)

\int_{0}^{4} (16 - x^{2}) d x

\frac{128}{3}

Solution steps

\int_{0}^{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_{0}^{4} 16 d x - \int_{0}^{4} x^{2} d x

\int_{0}^{4} 16 d x = 64

\int_{0}^{4} x^{2} d x = \frac{64}{3}

= 64 - \frac{64}{3}

Simplify

64 - \frac{64}{3} : \frac{128}{3}

= \frac{128}{3}

\frac{d}{d y} (\frac{x ^{2}}{y ^{2} + 1})

f (x) = 4 e^{x} cos (x), at x = 0

y = \frac{5 x ^{2} + 4}{x + 3}

\int_{0}^{\frac{π}{4}} x sin (8 x) d x

x^{2} ln (9 + x)

What is the integral from 0 to 4 of (16-x^2) ?
The integral from 0 to 4 of (16-x^2) is 128/3