What is the integral from 0 to 81 of 1/(sqrt(81-x)) ?

The integral from 0 to 81 of 1/(sqrt(81-x)) is 18

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integral from 0 to 81 of 1/(sqrt(81-x))

\int_{0}^{81} \frac{1}{81 - x} d x

18

Solution steps

\int_{0}^{81} \frac{1}{81 - x} d x

Apply u-substitution

: \int_{81}^{0} - \frac{1}{u} d u

= \int_{81}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{81} - \frac{1}{u} d u

Take the constant out:

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

= - (- \int_{0}^{81} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{81}

= - (- [2 u^{\frac{1}{2}}]_{0}^{81})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{81}) : [2 u]_{0}^{81}

= [2 u]_{0}^{81}

Compute the boundaries:

18

= 18

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\frac{d}{d x} (\frac{4 x}{x - 3})

x y^{2} y^{^{'}} = y^{3} - 3 x^{3}, y (1) = 3

What is the integral from 0 to 81 of 1/(sqrt(81-x)) ?
The integral from 0 to 81 of 1/(sqrt(81-x)) is 18