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_{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}

= - (- [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

y = 5 x, x = 5, y = \frac{5}{x}

\frac{d}{d x} (x^{3} + tan (sin (x)) + 2)

\frac{d}{d x} (- x^{2} + 1.8 x + 2.5)

y^{^{''}} - y = 0, y (0) = 1, y^{^{'}} (0) = 1

x \to 0 - lim (ln ∣ e^{x} - 1 ∣)

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