What is the integral from 0 to 1 of 4x^3(8-x^4) ?

The integral from 0 to 1 of 4x^3(8-x^4) is 15/2

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integral from 0 to 1 of 4x^3(8-x^4)

\int_{0}^{1} 4 x^{3} (8 - x^{4}) d x

\frac{15}{2}

Decimal

7.5

Solution steps

\int_{0}^{1} 4 x^{3} (8 - x^{4}) d x

Take the constant out:

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

= 4 \cdot \int_{0}^{1} x^{3} (8 - x^{4}) d x

Expand

x^{3} (8 - x^{4}) : 8 x^{3} - x^{7}

= 4 \cdot \int_{0}^{1} 8 x^{3} - x^{7} 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

= 4 (\int_{0}^{1} 8 x^{3} d x - \int_{0}^{1} x^{7} d x)

\int_{0}^{1} 8 x^{3} d x = 2

\int_{0}^{1} x^{7} d x = \frac{1}{8}

= 4 (2 - \frac{1}{8})

Simplify

4 (2 - \frac{1}{8}) : \frac{15}{2}

= \frac{15}{2}

What is the integral from 0 to 1 of 4x^3(8-x^4) ?
The integral from 0 to 1 of 4x^3(8-x^4) is 15/2