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

The integral from 0 to 1 of 2x(x^2+1/2)^2 is 13/12

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

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

\frac{13}{12}

Decimal

1.08333 \dots

Solution steps

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

Take the constant out:

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

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

Multiply by the conjugate of

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

= 2 \cdot \int_{0}^{1} x (\frac{x ^{4} - \frac{1}{4}}{x ^{2} - \frac{1}{2}})^{2} d x

Find undefined (singularity) points:

x = \frac{1}{2}

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

= 2 (\int_{0}^{\frac{1}{2}} x (\frac{x ^{4} - \frac{1}{4}}{x ^{2} - \frac{1}{2}})^{2} d x + \int_{\frac{1}{2}}^{1} x (\frac{x ^{4} - \frac{1}{4}}{x ^{2} - \frac{1}{2}})^{2} d x)

\int_{0}^{\frac{1}{2}} x (\frac{x ^{4} - \frac{1}{4}}{x ^{2} - \frac{1}{2}})^{2} d x = \frac{7}{48}

\int_{\frac{1}{2}}^{1} x (\frac{x ^{4} - \frac{1}{4}}{x ^{2} - \frac{1}{2}})^{2} d x = \frac{19}{48}

= 2 (\frac{7}{48} + \frac{19}{48})

Simplify

2 (\frac{7}{48} + \frac{19}{48}) : \frac{13}{12}

= \frac{13}{12}

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What is the integral from 0 to 1 of 2x(x^2+1/2)^2 ?
The integral from 0 to 1 of 2x(x^2+1/2)^2 is 13/12