What is the integral from sqrt(y) to 2 of 3/16 xy ?

The integral from sqrt(y) to 2 of 3/16 xy is (3y(4-y))/(32)

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integral from sqrt(y) to 2 of 3/16 xy

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

\frac{3 y ( 4 - y )}{32}

Solution steps

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

Take the constant out:

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

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

Multiply fractions:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

2 - \frac{y}{2}

= \frac{3 y}{16} (2 - \frac{y}{2})

Simplify

= \frac{3 y ( 4 - y )}{32}

x^{2} + 21, at x = 2

x \to 1 lim (\frac{4 x - 4}{x - 1})

\int 16 x sin (x) d x

\frac{1 + x}{1 - x}

\int_{0}^{e} ln (x) d x

What is the integral from sqrt(y) to 2 of 3/16 xy ?
The integral from sqrt(y) to 2 of 3/16 xy is (3y(4-y))/(32)