What is the integral from-1 to 1 of 2y^2 ?

The integral from-1 to 1 of 2y^2 is 4/3

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integral from-1 to 1 of 2y^2

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

\frac{4}{3}

Decimal Notation

1.33333 \dots

Solution steps

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

Take the constant out:

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

= 2 \cdot \int_{- 1}^{1} y^{2} d y

Apply the Power Rule

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

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

Compute the boundaries:

\frac{2}{3}

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

Simplify

= \frac{4}{3}

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