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

The integral from 0 to 1 of x^2(1+x^3)^2 is 7/9

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

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

\frac{7}{9}

Decimal

0.77777 \dots

Solution steps

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

Expand

x^{2} (1 + x^{3})^{2} : x^{2} + 2 x^{5} + x^{8}

= \int_{0}^{1} x^{2} + 2 x^{5} + x^{8} 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^{2} d x + \int_{0}^{1} 2 x^{5} d x + \int_{0}^{1} x^{8} d x

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

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

\int_{0}^{1} x^{8} d x = \frac{1}{9}

= \frac{1}{3} + \frac{1}{3} + \frac{1}{9}

Simplify

\frac{1}{3} + \frac{1}{3} + \frac{1}{9} : \frac{7}{9}

= \frac{7}{9}

What is the integral from 0 to 1 of x^2(1+x^3)^2 ?
The integral from 0 to 1 of x^2(1+x^3)^2 is 7/9