What is the integral from-1 to 1 of [(2-x^2)-x^2] ?

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

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integral from-1 to 1 of [(2-x^2)-x^2]

\int_{- 1}^{1} [(2 - x^{2}) - x^{2}] d x

\frac{8}{3}

Decimal

2.66666 \dots

Solution steps

\int_{- 1}^{1} 2 - x^{2} - x^{2} d x

Simplify

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

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

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

= 4 - \frac{4}{3}

Simplify

4 - \frac{4}{3} : \frac{8}{3}

= \frac{8}{3}

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