What is the integral from 0 to y of (2x^2)/(y^2) ?

The integral from 0 to y of (2x^2)/(y^2) is (2y)/3

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integral from 0 to y of (2x^2)/(y^2)

\int_{0}^{y} \frac{2 x ^{2}}{y ^{2}} d x

\frac{2 y}{3}

Solution steps

\int_{0}^{y} \frac{2 x ^{2}}{y ^{2}} d x

Take the constant out:

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

= \frac{2}{y ^{2}} \cdot \int_{0}^{y} x^{2} d x

Apply the Power Rule

: [\frac{x ^{3}}{3}]_{0}^{y}

= \frac{2}{y ^{2}} [\frac{x ^{3}}{3}]_{0}^{y}

Compute the boundaries:

\frac{y ^{3}}{3}

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

Simplify

= \frac{2 y}{3}

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What is the integral from 0 to y of (2x^2)/(y^2) ?
The integral from 0 to y of (2x^2)/(y^2) is (2y)/3