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

The integral from 0 to 4 of 2x(x^2+4)^2 is 7936/3

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

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

\frac{7936}{3}

Decimal Notation

2645.33333 \dots

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{0}^{4} x (x^{2} + 4)^{2} d x

Expand

x (x^{2} + 4)^{2} : x^{5} + 8 x^{3} + 16 x

= 2 \cdot \int_{0}^{4} x^{5} + 8 x^{3} + 16 x 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

= 2 (\int_{0}^{4} x^{5} d x + \int_{0}^{4} 8 x^{3} d x + \int_{0}^{4} 16 x d x)

\int_{0}^{4} x^{5} d x = \frac{2048}{3}

\int_{0}^{4} 8 x^{3} d x = 512

\int_{0}^{4} 16 x d x = 128

= 2 (\frac{2048}{3} + 512 + 128)

Simplify

2 (\frac{2048}{3} + 512 + 128) : \frac{7936}{3}

= \frac{7936}{3}

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What is the integral from 0 to 4 of 2x(x^2+4)^2 ?
The integral from 0 to 4 of 2x(x^2+4)^2 is 7936/3