What is the integral from 0 to 5 of 1/((5-x)^{1/3)} ?

The integral from 0 to 5 of 1/((5-x)^{1/3)} is (3*5^{2/3})/2

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integral from 0 to 5 of 1/((5-x)^{1/3)}

\int_{0}^{5} \frac{1}{( 5 - x ) ^{\frac{1}{3}}} d x

\frac{3 \cdot 5 ^{\frac{2}{3}}}{2}

Decimal

4.38602 \dots

Solution steps

\int_{0}^{5} \frac{1}{( 5 - x ) ^{\frac{1}{3}}} d x

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{( 5 - x ) ^{\frac{1}{3}}} = (5 - x)^{- \frac{1}{3}}

= \int_{0}^{5} (5 - x)^{- \frac{1}{3}} d x

Apply u-substitution

: \int_{5}^{0} - \frac{1}{u ^{\frac{1}{3}}} d u

= \int_{5}^{0} - \frac{1}{u ^{\frac{1}{3}}} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{5} - \frac{1}{u ^{\frac{1}{3}}} d u

Take the constant out:

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

= - (- \int_{0}^{5} \frac{1}{u ^{\frac{1}{3}}} d u)

Apply the Power Rule

: [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{5}

= - (- [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{5})

Simplify

= [\frac{3}{2} u^{\frac{2}{3}}]_{0}^{5}

Compute the boundaries:

5^{\frac{2}{3}} \cdot \frac{3}{2}

= 5^{\frac{2}{3}} \cdot \frac{3}{2}

Simplify

= \frac{3 \cdot 5 ^{\frac{2}{3}}}{2}

3 x^{4} - 7 ln (x)

\int \frac{x ^{4} + 2 x ^{3} - 3}{x ^{3}} d x

\frac{1}{( 1 - x ) ^{2}}

5 x y (\frac{d y}{d x}) = 1

\frac{d}{d x} (3 - 7 x - x^{2})

What is the integral from 0 to 5 of 1/((5-x)^{1/3)} ?
The integral from 0 to 5 of 1/((5-x)^{1/3)} is (3*5^{2/3})/2