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

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

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

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

\frac{1}{3}

Solution steps

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

Take the constant out:

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

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

Apply the Power Rule

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

= \frac{1}{2} [\frac{x ^{3}}{3}]_{- 1}^{1}

Compute the boundaries:

\frac{2}{3}

= \frac{1}{2} \cdot \frac{2}{3}

Simplify

= \frac{1}{3}

\int_{7}^{14} \frac{6}{49} x d x

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\int_{- 1}^{0} 4 x^{2} 4 x^{3} + 6 d x

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

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