What is the integral from-11 to 0 of 1/(sqrt(25-x)) ?

The integral from-11 to 0 of 1/(sqrt(25-x)) is 2

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integral from-11 to 0 of 1/(sqrt(25-x))

\int_{- 11}^{0} \frac{1}{25 - x} d x

2

Solution steps

\int_{- 11}^{0} \frac{1}{25 - x} d x

Apply u-substitution

= \int_{36}^{25} - \frac{1}{u} d u

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

= - \int_{25}^{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_{25}^{36} \frac{1}{u} d u)

Apply the Power Rule

= - (- [2 u^{\frac{1}{2}}]_{25}^{36})

Simplify

- (- [2 u^{\frac{1}{2}}]_{25}^{36}) : [2 u]_{25}^{36}

= [2 u]_{25}^{36}

Compute the boundaries:

2

= 2

\int 12 t d t

2^{(5 x)}

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

- \frac{1}{s ^{2}} e^{- 5 s} + \frac{1}{s ^{2}}

\int \frac{3 x ^{2} - 8 x - 9}{x ^{3} - 4 x ^{2} - 9 x + 36} d x

What is the integral from-11 to 0 of 1/(sqrt(25-x)) ?
The integral from-11 to 0 of 1/(sqrt(25-x)) is 2