What is the integral from-7 to 0 of 1/(sqrt(9-x)) ?

The integral from-7 to 0 of 1/(sqrt(9-x)) is 2

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integral from-7 to 0 of 1/(sqrt(9-x))

\int_{- 7}^{0} \frac{1}{9 - x} d x

2

Solution steps

\int_{- 7}^{0} \frac{1}{9 - x} d x

Apply u-substitution

: \int_{16}^{9} - \frac{1}{u} d u

= \int_{16}^{9} - \frac{1}{u} d u

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

= - \int_{9}^{16} - \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}^{16} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{16}

= - (- [2 u^{\frac{1}{2}}]_{9}^{16})

Simplify

- (- [2 u^{\frac{1}{2}}]_{9}^{16}) : [2 u]_{9}^{16}

= [2 u]_{9}^{16}

Compute the boundaries:

2

= 2

\frac{d}{d x} (\frac{x ^{5}}{x ^{3}})

\frac{d}{d x} (5 x^{2} cos (x) + x)

\frac{\partial}{\partial y} (\frac{( y + x )}{( y - x )})

f (x) = x^{2} - 2

2 \frac{d y}{d x} = x y (x + 1)

What is the integral from-7 to 0 of 1/(sqrt(9-x)) ?
The integral from-7 to 0 of 1/(sqrt(9-x)) is 2