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

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

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

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

\frac{8}{45}

Decimal

0.17777 \dots

Solution steps

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

Expand

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

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

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

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

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

\int_{- 1}^{1} \frac{1}{9} d t = \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}

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