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

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

Popular >

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

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

\frac{2048}{15}

Decimal Notation

136.53333 \dots

Solution steps

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

Simplify

(4 x)^{2} - (x^{2})^{2} : 16 x^{2} - x^{4}

= \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 y} (ln (3 x + 5 y))

\frac{d}{d x} (5 e^{- 5 x})

n = 0 \sum \infty \frac{2}{4 n ^{2} - 1}

\int \frac{x + 2}{x ^{2} + 1} d x

y = 1 - x^{2} arcsin (x)

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