What is the integral from 0 to 36 of 1/(sqrt(36-x)) ?

The integral from 0 to 36 of 1/(sqrt(36-x)) is 12

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integral from 0 to 36 of 1/(sqrt(36-x))

\int_{0}^{36} \frac{1}{36 - x} d x

12

Solution steps

\int_{0}^{36} \frac{1}{36 - x} d x

Apply u-substitution

= \int_{36}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{36} - \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}^{36} \frac{1}{u} d u)

Apply the Power Rule

= - (- [2 u^{\frac{1}{2}}]_{0}^{36})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{36}) : [2 u]_{0}^{36}

= [2 u]_{0}^{36}

Compute the boundaries:

12

= 12

What is the integral from 0 to 36 of 1/(sqrt(36-x)) ?
The integral from 0 to 36 of 1/(sqrt(36-x)) is 12