What is the integral from 0 to 5 of 1/(sqrt(5-x)) ?

The integral from 0 to 5 of 1/(sqrt(5-x)) is 2sqrt(5)

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integral from 0 to 5 of 1/(sqrt(5-x))

\int_{0}^{5} \frac{1}{5 - x} d x

25

Solution steps

\int_{0}^{5} \frac{1}{5 - x} d x

Apply u-substitution

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

= \int_{5}^{0} - \frac{1}{u} d u

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

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

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{5}

= - (- [2 u^{\frac{1}{2}}]_{0}^{5})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{5}) : [2 u]_{0}^{5}

= [2 u]_{0}^{5}

Compute the boundaries:

25

= 25

\int_{0}^{5} - x^{2} + 9 d x

\int_{0}^{b} \frac{1}{32} v e^{- \frac{v ^{2}}{64}} d v

\int_{0.25}^{4} \frac{1}{x} d x

\int_{1}^{2} 4^{5 x} d x

\int_{- 5}^{0} 2 (t + 1)^{\frac{1}{3}} d t

What is the integral from 0 to 5 of 1/(sqrt(5-x)) ?
The integral from 0 to 5 of 1/(sqrt(5-x)) is 2sqrt(5)