What is the integral from-5 to 0 of 1/(sqrt(4-x)) ?

The integral from-5 to 0 of 1/(sqrt(4-x)) is 2

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integral from-5 to 0 of 1/(sqrt(4-x))

\int_{- 5}^{0} \frac{1}{4 - x} d x

2

Solution steps

\int_{- 5}^{0} \frac{1}{4 - x} d x

Apply u-substitution

: \int_{9}^{4} - \frac{1}{u} d u

= \int_{9}^{4} - \frac{1}{u} d u

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

= - \int_{4}^{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_{4}^{9} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{4}^{9}

= - (- [2 u^{\frac{1}{2}}]_{4}^{9})

Simplify

- (- [2 u^{\frac{1}{2}}]_{4}^{9}) : [2 u]_{4}^{9}

= [2 u]_{4}^{9}

Compute the boundaries:

2

= 2

\int (\frac{1}{x ^{2}} - \frac{4}{x ^{3}}) d x

\frac{\partial}{\partial x} (3 x z + y^{2})

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

f (x) = tan (3 x), at x = \frac{π}{9}

\int 2 + 2 cos (4 x) d x

What is the integral from-5 to 0 of 1/(sqrt(4-x)) ?
The integral from-5 to 0 of 1/(sqrt(4-x)) is 2