What is the integral from 1 to 5 of 1/96 xy ?

The integral from 1 to 5 of 1/96 xy is x/8

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integral from 1 to 5 of 1/96 xy

\int_{1}^{5} \frac{1}{96} x y d y

\frac{x}{8}

Solution steps

\int_{1}^{5} \frac{1}{96} x y d y

Take the constant out:

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

= \frac{1}{96} x \cdot \int_{1}^{5} y d y

Simplify

\frac{1}{96} x : \frac{x}{96}

= \frac{x}{96} \cdot \int_{1}^{5} y d y

Apply the Power Rule

: [\frac{y ^{2}}{2}]_{1}^{5}

= \frac{x}{96} [\frac{y ^{2}}{2}]_{1}^{5}

Compute the boundaries:

12

= \frac{x}{96} \cdot 12

Simplify

= \frac{x}{8}

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What is the integral from 1 to 5 of 1/96 xy ?
The integral from 1 to 5 of 1/96 xy is x/8