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

The integral from-2 to 2 of x^2 is 16/3

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integral from-2 to 2 of x^2

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

\frac{16}{3}

Decimal Notation

5.33333 \dots

Solution steps

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

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{- 2}^{2}

= [\frac{x ^{3}}{3}]_{- 2}^{2}

Compute the boundaries:

\frac{16}{3}

= \frac{16}{3}

\frac{b}{( 1 - ( \frac{y}{x} ) )}

x^{\frac{2}{3}} - x^{\frac{1}{3}}

\int 7 sin (2) (x) cos (3) (x) d x

\frac{d}{d x} (\frac{x ^{2} ( x - 1 )}{( x + 1 ) ^{2}})

\frac{d ^{2}}{d x ^{2}} (\frac{7}{8 t ^{6}})

What is the integral from-2 to 2 of x^2 ?
The integral from-2 to 2 of x^2 is 16/3