What is the integral from 0 to 1 of 6(y/x)^4 ?

The integral from 0 to 1 of 6(y/x)^4 is 6/(5x^4)

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integral from 0 to 1 of 6(y/x)^4

\int_{0}^{1} 6 (\frac{y}{x})^{4} d y

\frac{6}{5 x ^{4}}

Solution steps

\int_{0}^{1} 6 (\frac{y}{x})^{4} d y

Take the constant out:

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

= 6 \cdot \int_{0}^{1} (\frac{y}{x})^{4} d y

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= 6 \cdot \int_{0}^{1} \frac{y ^{4}}{x ^{4}} d y

Take the constant out:

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

= 6 \cdot \frac{1}{x ^{4}} \cdot \int_{0}^{1} y^{4} d y

Apply the Power Rule

: [\frac{y ^{5}}{5}]_{0}^{1}

= 6 \cdot \frac{1}{x ^{4}} [\frac{y ^{5}}{5}]_{0}^{1}

Simplify

6 \cdot \frac{1}{x ^{4}} [\frac{y ^{5}}{5}]_{0}^{1} : \frac{6 [ \frac{y ^{5}}{5} ] _{0}^{1}}{x ^{4}}

= \frac{6 [ \frac{y ^{5}}{5} ] _{0}^{1}}{x ^{4}}

Compute the boundaries:

\frac{1}{5}

= \frac{6 \cdot \frac{1}{5}}{x ^{4}}

Simplify

= \frac{6}{5 x ^{4}}

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What is the integral from 0 to 1 of 6(y/x)^4 ?
The integral from 0 to 1 of 6(y/x)^4 is 6/(5x^4)