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

The integral from 0 to 2 of 5/((2-x)^{2/3)} is 15*2^{1/3}

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

\int_{0}^{2} \frac{5}{( 2 - x ) ^{\frac{2}{3}}} d x

15 \cdot 2^{\frac{1}{3}}

Decimal Notation

18.89881 \dots

Solution steps

\int_{0}^{2} \frac{5}{( 2 - x ) ^{\frac{2}{3}}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{2} \frac{1}{( 2 - x ) ^{\frac{2}{3}}} d x

Apply u-substitution

: \int_{2}^{0} - \frac{1}{u ^{\frac{2}{3}}} d u

= 5 \cdot \int_{2}^{0} - \frac{1}{u ^{\frac{2}{3}}} d u

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

= 5 (- \int_{0}^{2} - \frac{1}{u ^{\frac{2}{3}}} d u)

Take the constant out:

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

= 5 (- (- \int_{0}^{2} \frac{1}{u ^{\frac{2}{3}}} d u))

Apply the Power Rule

: [3 u^{\frac{1}{3}}]_{0}^{2}

= 5 (- (- [3 u^{\frac{1}{3}}]_{0}^{2}))

Simplify

= 5 [3 u^{\frac{1}{3}}]_{0}^{2}

Compute the boundaries:

3 \cdot 2^{\frac{1}{3}}

= 5 \cdot 3 \cdot 2^{\frac{1}{3}}

Simplify

= 15 \cdot 2^{\frac{1}{3}}

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What is the integral from 0 to 2 of 5/((2-x)^{2/3)} ?
The integral from 0 to 2 of 5/((2-x)^{2/3)} is 15*2^{1/3}