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

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

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

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

\frac{4}{3}

Decimal

1.33333 \dots

Solution steps

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

Take the constant out:

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

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

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

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

= - (\frac{2}{3} - 2)

Simplify

- (\frac{2}{3} - 2) : \frac{4}{3}

= \frac{4}{3}

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