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

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

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

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

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

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

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{- 1}^{1}

= [\frac{x ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{2}{3}

\frac{d}{d x} (\frac{e ^{3 x}}{x ^{12}})

\frac{d}{d x} (\frac{2}{( 3 x ) ^{2}})

x \to 0.1 lim (\frac{5 sin ( x )}{x})

x \to 6 lim (\frac{x ^{2} - 3 x - 18}{x - 6})

(- 3.2) (6.3)

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