What is the integral from 0 to 2 of 3/32 x^2y^3 ?

The integral from 0 to 2 of 3/32 x^2y^3 is (3x^2)/8

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integral from 0 to 2 of 3/32 x^2y^3

\int_{0}^{2} \frac{3}{32} x^{2} y^{3} d y

\frac{3 x ^{2}}{8}

Solution steps

\int_{0}^{2} \frac{3}{32} x^{2} y^{3} d y

Take the constant out:

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

= \frac{3}{32} x^{2} \cdot \int_{0}^{2} y^{3} d y

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 x ^{2}}{32} \cdot \int_{0}^{2} y^{3} d y

Apply the Power Rule

: [\frac{y ^{4}}{4}]_{0}^{2}

= \frac{3 x ^{2}}{32} [\frac{y ^{4}}{4}]_{0}^{2}

Compute the boundaries:

4

= \frac{3 x ^{2}}{32} \cdot 4

Simplify

= \frac{3 x ^{2}}{8}

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What is the integral from 0 to 2 of 3/32 x^2y^3 ?
The integral from 0 to 2 of 3/32 x^2y^3 is (3x^2)/8