What is the integral from 0 to 1 of x^3(1+x)^2 ?

The integral from 0 to 1 of x^3(1+x)^2 is 49/60

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integral from 0 to 1 of x^3(1+x)^2

\int_{0}^{1} x^{3} (1 + x)^{2} d x

\frac{49}{60}

Decimal Notation

0.81666 \dots

Solution steps

\int_{0}^{1} x^{3} (1 + x)^{2} d x

Expand

x^{3} (1 + x)^{2} : x^{3} + 2 x^{4} + x^{5}

= \int_{0}^{1} x^{3} + 2 x^{4} + x^{5} 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}^{1} x^{3} d x + \int_{0}^{1} 2 x^{4} d x + \int_{0}^{1} x^{5} d x

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

\int_{0}^{1} 2 x^{4} d x = \frac{2}{5}

\int_{0}^{1} x^{5} d x = \frac{1}{6}

= \frac{1}{4} + \frac{2}{5} + \frac{1}{6}

Simplify

\frac{1}{4} + \frac{2}{5} + \frac{1}{6} : \frac{49}{60}

= \frac{49}{60}

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What is the integral from 0 to 1 of x^3(1+x)^2 ?
The integral from 0 to 1 of x^3(1+x)^2 is 49/60