What is the integral from 0 to 1 of 7x^3(-x^4+1)^4 ?

The integral from 0 to 1 of 7x^3(-x^4+1)^4 is 7/20

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integral from 0 to 1 of 7x^3(-x^4+1)^4

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

\frac{7}{20}

Decimal

0.35

Solution steps

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

Take the constant out:

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

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

Expand

x^{3} (- x^{4} + 1)^{4} : x^{19} - 4 x^{15} + 6 x^{11} - 4 x^{7} + x^{3}

= 7 \cdot \int_{0}^{1} x^{19} - 4 x^{15} + 6 x^{11} - 4 x^{7} + x^{3} 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

= 7 (\int_{0}^{1} x^{19} d x - \int_{0}^{1} 4 x^{15} d x + \int_{0}^{1} 6 x^{11} d x - \int_{0}^{1} 4 x^{7} d x + \int_{0}^{1} x^{3} d x)

\int_{0}^{1} x^{19} d x = \frac{1}{20}

\int_{0}^{1} 4 x^{15} d x = \frac{1}{4}

\int_{0}^{1} 6 x^{11} d x = \frac{1}{2}

\int_{0}^{1} 4 x^{7} d x = \frac{1}{2}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 7 (\frac{1}{20} - \frac{1}{4} + \frac{1}{2} - \frac{1}{2} + \frac{1}{4})

Simplify

7 (\frac{1}{20} - \frac{1}{4} + \frac{1}{2} - \frac{1}{2} + \frac{1}{4}) : \frac{7}{20}

= \frac{7}{20}

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What is the integral from 0 to 1 of 7x^3(-x^4+1)^4 ?
The integral from 0 to 1 of 7x^3(-x^4+1)^4 is 7/20