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

The integral from-1 to 1 of (x^2-1/3)^2 is 8/45

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

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

\frac{8}{45}

Decimal Notation

0.17777 \dots

Solution steps

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

Expand

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

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

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

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

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

= \frac{2}{5} - \frac{4}{9} + \frac{2}{9}

Simplify

\frac{2}{5} - \frac{4}{9} + \frac{2}{9} : \frac{8}{45}

= \frac{8}{45}

\int \frac{5 x ^{2}}{( 3 + x ^{3} ) ^{2}} d x

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\frac{\partial}{\partial y} (\frac{( x - y )}{( x + y )})

\int 4 x ln (5 x) d x

y^{^{'}} = 4 y^{2} + x y^{2}, y (0) = 1

What is the integral from-1 to 1 of (x^2-1/3)^2 ?
The integral from-1 to 1 of (x^2-1/3)^2 is 8/45