What is the integral from 0 to x^2 of 3/16 xy ?

The integral from 0 to x^2 of 3/16 xy is (3x^5)/(32)

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integral from 0 to x^2 of 3/16 xy

\int_{0}^{x^{2}} \frac{3}{16} x y d y

\frac{3 x ^{5}}{32}

Solution steps

\int_{0}^{x^{2}} \frac{3}{16} x y d y

Take the constant out:

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

= \frac{3}{16} x \cdot \int_{0}^{x^{2}} y d y

Multiply fractions:

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

= \frac{3 x}{16} \cdot \int_{0}^{x^{2}} y d y

Apply the Power Rule

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

= \frac{3 x}{16} [\frac{y ^{2}}{2}]_{0}^{x^{2}}

Compute the boundaries:

\frac{x ^{4}}{2}

= \frac{3 x}{16} \cdot \frac{x ^{4}}{2}

Simplify

= \frac{3 x ^{5}}{32}

\int (2 x - 1) e^{x (x - 1)} d x

x \to 1 + lim (x^{\frac{7}{1 - x}})

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x y^{^{'}} + 3 y = x^{3} - x

\int x^{- 0.3} d x

What is the integral from 0 to x^2 of 3/16 xy ?
The integral from 0 to x^2 of 3/16 xy is (3x^5)/(32)