What is the integral from 0 to 49 of 1/(sqrt(49-x)) ?

The integral from 0 to 49 of 1/(sqrt(49-x)) is 14

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integral from 0 to 49 of 1/(sqrt(49-x))

\int_{0}^{49} \frac{1}{49 - x} d x

14

Solution steps

\int_{0}^{49} \frac{1}{49 - x} d x

Apply u-substitution

: \int_{49}^{0} - \frac{1}{u} d u

= \int_{49}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{49} - \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}^{49} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{49}

= - (- [2 u^{\frac{1}{2}}]_{0}^{49})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{49}) : [2 u]_{0}^{49}

= [2 u]_{0}^{49}

Compute the boundaries:

14

= 14

x sin (\frac{5}{x})

\frac{d}{d x} (\frac{x}{1 - ln ( x - 9 )})

x y^{^{'}} - y = \frac{5}{4} x ln (x)

(6 - x) x + 15

5 e^{- 3 t}

What is the integral from 0 to 49 of 1/(sqrt(49-x)) ?
The integral from 0 to 49 of 1/(sqrt(49-x)) is 14