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

The integral from 0 to 4 of 16x^2-x^4 is 2048/15

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

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

\frac{2048}{15}

Decimal Notation

136.53333 \dots

Solution steps

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

\int_{0}^{4} 16 x^{2} d x = \frac{1024}{3}

\int_{0}^{4} x^{4} d x = \frac{1024}{5}

= \frac{1024}{3} - \frac{1024}{5}

Simplify

\frac{1024}{3} - \frac{1024}{5} : \frac{2048}{15}

= \frac{2048}{15}

\frac{\partial}{\partial x} (x^{2} sin (6 t))

\frac{d}{d x} (- \frac{e ^{- x} ( x ^{3} + 3 ( x ^{2} + 2 x + 2 ) )}{x ^{4}})

f (x) = \frac{x ^{2} + 8}{x + 28}, (5, 1)

(4, - 9), (2, - 2)

\int x^{m - 1} d x

What is the integral from 0 to 4 of 16x^2-x^4 ?
The integral from 0 to 4 of 16x^2-x^4 is 2048/15