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

The integral from 0 to 1 of x^2(1/2+3/2 x^2) is 7/15

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

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

\frac{7}{15}

Decimal Notation

0.46666 \dots

Solution steps

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

Multiply by the conjugate of

\frac{1}{2} + \frac{3}{2} x^{2} : \frac{\frac{1}{4} - \frac{9}{4} x ^{4}}{\frac{1}{2} - \frac{3}{2} x ^{2}}

= \int_{0}^{1} x^{2} \frac{\frac{1}{4} - \frac{9}{4} x ^{4}}{\frac{1}{2} - \frac{3}{2} x ^{2}} d x

Find undefined (singularity) points:

x = \frac{1}{3}

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{\frac{1}{3}} x^{2} \frac{\frac{1}{4} - \frac{9}{4} x ^{4}}{\frac{1}{2} - \frac{3}{2} x ^{2}} d x + \int_{\frac{1}{3}}^{1} x^{2} \frac{\frac{1}{4} - \frac{9}{4} x ^{4}}{\frac{1}{2} - \frac{3}{2} x ^{2}} d x

\int_{0}^{\frac{1}{3}} x^{2} \frac{\frac{1}{4} - \frac{9}{4} x ^{4}}{\frac{1}{2} - \frac{3}{2} x ^{2}} d x = \frac{4}{45 3}

\int_{\frac{1}{3}}^{1} x^{2} \frac{\frac{1}{4} - \frac{9}{4} x ^{4}}{\frac{1}{2} - \frac{3}{2} x ^{2}} d x = \frac{21 3 - 4}{45 3}

= \frac{4}{45 3} + \frac{21 3 - 4}{45 3}

Simplify

\frac{4}{45 3} + \frac{21 3 - 4}{45 3} : \frac{7}{15}

= \frac{7}{15}

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What is the integral from 0 to 1 of x^2(1/2+3/2 x^2) ?
The integral from 0 to 1 of x^2(1/2+3/2 x^2) is 7/15