What is the integral from 0 to 2 of 1/((2-x)^{4/5)} ?

The integral from 0 to 2 of 1/((2-x)^{4/5)} is 5*2^{1/5}

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integral from 0 to 2 of 1/((2-x)^{4/5)}

\int_{0}^{2} \frac{1}{( 2 - x ) ^{\frac{4}{5}}} d x

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

Decimal

5.74349 \dots

Solution steps

\int_{0}^{2} \frac{1}{( 2 - x ) ^{\frac{4}{5}}} d x

Apply u-substitution

: \int_{2}^{0} - \frac{1}{u ^{\frac{4}{5}}} d u

= \int_{2}^{0} - \frac{1}{u ^{\frac{4}{5}}} d u

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

= - \int_{0}^{2} - \frac{1}{u ^{\frac{4}{5}}} d u

Take the constant out:

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

= - (- \int_{0}^{2} \frac{1}{u ^{\frac{4}{5}}} d u)

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

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

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

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What is the integral from 0 to 2 of 1/((2-x)^{4/5)} ?
The integral from 0 to 2 of 1/((2-x)^{4/5)} is 5*2^{1/5}