What is the integral from 1 to 6 of 1/((6-x)^{2/3)} ?

The integral from 1 to 6 of 1/((6-x)^{2/3)} is 3*5^{1/3}

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integral from 1 to 6 of 1/((6-x)^{2/3)}

\int_{1}^{6} \frac{1}{( 6 - x ) ^{\frac{2}{3}}} d x

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

Decimal Notation

5.12992 \dots

Solution steps

\int_{1}^{6} \frac{1}{( 6 - x ) ^{\frac{2}{3}}} d x

Apply u-substitution

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

= \int_{5}^{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

= - \int_{0}^{5} - \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

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

Apply the Power Rule

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

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

Simplify

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

Compute the boundaries:

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

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

\int \frac{3}{2} (t + 1)^{\frac{1}{2}} d t

\frac{1}{t} \frac{d y}{d t} - \frac{1}{1 + t ^{2}} = 0

\frac{x ^{3} + 7}{x}

\int x^{2} \cdot ln (2 x - 3) d x

\int 5 cos^{2} (x t a) n^{3} x d x

What is the integral from 1 to 6 of 1/((6-x)^{2/3)} ?
The integral from 1 to 6 of 1/((6-x)^{2/3)} is 3*5^{1/3}