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

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

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

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

\frac{2}{15}

Solution steps

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

Complete the square

3 x^{2} - 2 x : 3 (x - \frac{1}{3})^{2} - \frac{1}{3}

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

Apply u-substitution

: \int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} - 2 u^{2} + \frac{1}{9} d u

= \int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} - 2 u^{2} + \frac{1}{9} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} d u - \int_{- \frac{1}{3}}^{\frac{2}{3}} 2 u^{2} d u + \int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{9} d u

\int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} d u = \frac{11}{45}

\int_{- \frac{1}{3}}^{\frac{2}{3}} 2 u^{2} d u = \frac{2}{9}

\int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{9} d u = \frac{1}{9}

= \frac{11}{45} - \frac{2}{9} + \frac{1}{9}

Simplify

\frac{11}{45} - \frac{2}{9} + \frac{1}{9} : \frac{2}{15}

= \frac{2}{15}

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