What is the integral from-9 to 0 of 1/(sqrt(16-x)) ?

The integral from-9 to 0 of 1/(sqrt(16-x)) is 2

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integral from-9 to 0 of 1/(sqrt(16-x))

\int_{- 9}^{0} \frac{1}{16 - x} d x

2

Solution steps

\int_{- 9}^{0} \frac{1}{16 - x} d x

Apply u-substitution

: \int_{25}^{16} - \frac{1}{u} d u

= \int_{25}^{16} - \frac{1}{u} d u

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

= - \int_{16}^{25} - \frac{1}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{16}^{25} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{16}^{25}

= - (- [2 u^{\frac{1}{2}}]_{16}^{25})

Simplify

- (- [2 u^{\frac{1}{2}}]_{16}^{25}) : [2 u]_{16}^{25}

= [2 u]_{16}^{25}

Compute the boundaries:

2

= 2

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What is the integral from-9 to 0 of 1/(sqrt(16-x)) ?
The integral from-9 to 0 of 1/(sqrt(16-x)) is 2