What is the integral from-40 to 0 of 1/(sqrt(9-x)) ?

The integral from-40 to 0 of 1/(sqrt(9-x)) is 8

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integral from-40 to 0 of 1/(sqrt(9-x))

\int_{- 40}^{0} \frac{1}{9 - x} d x

8

Solution steps

\int_{- 40}^{0} \frac{1}{9 - x} d x

Apply u-substitution

: \int_{49}^{9} - \frac{1}{u} d u

= \int_{49}^{9} - \frac{1}{u} d u

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

= - \int_{9}^{49} - \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}^{49} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{49}

= - (- [2 u^{\frac{1}{2}}]_{9}^{49})

Simplify

- (- [2 u^{\frac{1}{2}}]_{9}^{49}) : [2 u]_{9}^{49}

= [2 u]_{9}^{49}

Compute the boundaries:

8

= 8

y = 30 e^{- 0.0000323 x}

\frac{1}{8} x, 8 x, \frac{2}{x}

\int_{2}^{8} \frac{5}{4 - 3 x} d x

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

(x^{2} + y) d x - (x) d y = 0

What is the integral from-40 to 0 of 1/(sqrt(9-x)) ?
The integral from-40 to 0 of 1/(sqrt(9-x)) is 8