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

The integral from 0 to 2 of (x^2-1)-(1-x^2) is 4/3

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integral from 0 to 2 of (x^2-1)-(1-x^2)

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Simplify

= \int_{0}^{2} 2 x^{2} - 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_{0}^{2} 2 x^{2} d x - \int_{0}^{2} 2 d x

\int_{0}^{2} 2 x^{2} d x = \frac{16}{3}

\int_{0}^{2} 2 d x = 4

= \frac{16}{3} - 4

Simplify

\frac{16}{3} - 4 : \frac{4}{3}

= \frac{4}{3}

What is the integral from 0 to 2 of (x^2-1)-(1-x^2) ?
The integral from 0 to 2 of (x^2-1)-(1-x^2) is 4/3