What is the integral from 0 to 9 of 1/(sqrt(9-x)) ?

The integral from 0 to 9 of 1/(sqrt(9-x)) is 6

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integral from 0 to 9 of 1/(sqrt(9-x))

\int_{0}^{9} \frac{1}{9 - x} d x

6

Solution steps

\int_{0}^{9} \frac{1}{9 - x} d x

Apply u-substitution

: \int_{9}^{0} - \frac{1}{u} d u

= \int_{9}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{9} - \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}^{9} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{9}

= - (- [2 u^{\frac{1}{2}}]_{0}^{9})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{9}) : [2 u]_{0}^{9}

= [2 u]_{0}^{9}

Compute the boundaries:

6

= 6

\frac{d}{d x} (4 x^{\frac{5}{4}} + 10 x^{\frac{3}{2}} + 10 x)

\frac{d}{d x} ((x - 4) \cdot a x^{2} + ln (\frac{x}{4}) \cdot b x^{4})

\int cos (8 t) d t

\int \frac{x ^{2} + 2 x - 1}{x ( x + 1 ) ( x - 1 )} d x

y^{^{'}} = 3 x + y, y (0) = 7

What is the integral from 0 to 9 of 1/(sqrt(9-x)) ?
The integral from 0 to 9 of 1/(sqrt(9-x)) is 6