What is the integral from-27 to 0 of 1/(sqrt(9-x)) ?

The integral from-27 to 0 of 1/(sqrt(9-x)) is 6

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integral from-27 to 0 of 1/(sqrt(9-x))

\int_{- 27}^{0} \frac{1}{9 - x} d x

6

Solution steps

\int_{- 27}^{0} \frac{1}{9 - x} d x

Apply u-substitution

: \int_{36}^{9} - \frac{1}{u} d u

= \int_{36}^{9} - \frac{1}{u} d u

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

= - \int_{9}^{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_{9}^{36} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{36}

= - (- [2 u^{\frac{1}{2}}]_{9}^{36})

Simplify

- (- [2 u^{\frac{1}{2}}]_{9}^{36}) : [2 u]_{9}^{36}

= [2 u]_{9}^{36}

Compute the boundaries:

6

= 6

x \to 0 lim ((\frac{1}{x})^{l n (x)})

x \to - 1 lim (- x^{2} + 5 x - 1)

\int \frac{1}{( x ^{2} + 100 ) ^{\frac{3}{2}}} d x

\frac{( s + 3 )}{s ^{2} + 6 s + 13}

\int \frac{13 x ^{3}}{9 + x ^{2}} d x

What is the integral from-27 to 0 of 1/(sqrt(9-x)) ?
The integral from-27 to 0 of 1/(sqrt(9-x)) is 6