What is the integral from 4 to 5 of 1/((y-4)^{1/3)} ?

The integral from 4 to 5 of 1/((y-4)^{1/3)} is 3/2

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integral from 4 to 5 of 1/((y-4)^{1/3)}

\int_{4}^{5} \frac{1}{( y - 4 ) ^{\frac{1}{3}}} d y

\frac{3}{2}

Decimal Notation

1.5

Solution steps

\int_{4}^{5} \frac{1}{( y - 4 ) ^{\frac{1}{3}}} d y

Apply Integration By Parts

: [\frac{y}{( y - 4 ) ^{\frac{1}{3}}} - \int - \frac{y}{3 ( y - 4 ) ^{\frac{4}{3}}} d y]_{4}^{5}

= [\frac{y}{( y - 4 ) ^{\frac{1}{3}}} - \int - \frac{y}{3 ( y - 4 ) ^{\frac{4}{3}}} d y]_{4}^{5}

\int - \frac{y}{3 ( y - 4 ) ^{\frac{4}{3}}} d y = - \frac{y - 12}{2 ( y - 4 ) ^{\frac{1}{3}}}

= [\frac{y}{( y - 4 ) ^{\frac{1}{3}}} - (- \frac{y - 12}{2 ( y - 4 ) ^{\frac{1}{3}}})]_{4}^{5}

Simplify

[\frac{y}{( y - 4 ) ^{\frac{1}{3}}} - (- \frac{y - 12}{2 ( y - 4 ) ^{\frac{1}{3}}})]_{4}^{5} : [\frac{3}{2} (y - 4)^{\frac{2}{3}}]_{4}^{5}

= [\frac{3}{2} (y - 4)^{\frac{2}{3}}]_{4}^{5}

Compute the boundaries:

\frac{3}{2}

= \frac{3}{2}

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What is the integral from 4 to 5 of 1/((y-4)^{1/3)} ?
The integral from 4 to 5 of 1/((y-4)^{1/3)} is 3/2