What is the integral from 0 to 25 of 1/(sqrt(25-x)) ?

The integral from 0 to 25 of 1/(sqrt(25-x)) is 10

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integral from 0 to 25 of 1/(sqrt(25-x))

\int_{0}^{25} \frac{1}{25 - x} d x

10

Solution steps

\int_{0}^{25} \frac{1}{25 - x} d x

Apply u-substitution

: \int_{25}^{0} - \frac{1}{u} d u

= \int_{25}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{25} - \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}^{25} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{25}

= - (- [2 u^{\frac{1}{2}}]_{0}^{25})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{25}) : [2 u]_{0}^{25}

= [2 u]_{0}^{25}

Compute the boundaries:

10

= 10

\int 12 x e^{6 x} d x

\frac{\partial}{\partial y} (x e^{y} + y)

\frac{d y}{d x} = x y^{\frac{1}{2}}

\frac{d}{d x} (2 ln (2))

\int \frac{( x ^{2} - 25 ) ^{\frac{3}{2}}}{x ^{6}} d x

What is the integral from 0 to 25 of 1/(sqrt(25-x)) ?
The integral from 0 to 25 of 1/(sqrt(25-x)) is 10