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

The integral from-8 to 0 of 1/(sqrt(1-x)) is 4

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integral from-8 to 0 of 1/(sqrt(1-x))

\int_{- 8}^{0} \frac{1}{1 - x} d x

4

Solution steps

\int_{- 8}^{0} \frac{1}{1 - x} d x

Apply u-substitution

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

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

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

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

Apply the Power Rule

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

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

Simplify

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

= [2 u]_{1}^{9}

Compute the boundaries:

4

= 4

\frac{d}{d z} (\frac{y}{x ^{2} + y ^{2} + z ^{2}})

\int 9 arctan (x) d x

\frac{d}{d x} (8718 e (587 x))

\frac{x ^{2}}{x + 8}

\frac{1}{5 - x}

What is the integral from-8 to 0 of 1/(sqrt(1-x)) ?
The integral from-8 to 0 of 1/(sqrt(1-x)) is 4