What is the integral from-20 to 0 of 1/(sqrt(16-x)) ?

The integral from-20 to 0 of 1/(sqrt(16-x)) is 4

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integral from-20 to 0 of 1/(sqrt(16-x))

\int_{- 20}^{0} \frac{1}{16 - x} d x

4

Solution steps

\int_{- 20}^{0} \frac{1}{16 - x} d x

Apply u-substitution

: \int_{36}^{16} - \frac{1}{u} d u

= \int_{36}^{16} - \frac{1}{u} d u

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

= - \int_{16}^{36} - \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}^{36} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{16}^{36}

= - (- [2 u^{\frac{1}{2}}]_{16}^{36})

Simplify

- (- [2 u^{\frac{1}{2}}]_{16}^{36}) : [2 u]_{16}^{36}

= [2 u]_{16}^{36}

Compute the boundaries:

4

= 4

\frac{e ^{- 2 s}}{( s ^{2} + 16 )}

y = 4 x, x = 4, y = \frac{4}{x}

\int_{0}^{1} (x^{3} - 1) d x

x \to - 3 lim (\frac{x ^{2}}{x + 3})

\int \frac{3 y}{2 y ^{2} + 5} d y

What is the integral from-20 to 0 of 1/(sqrt(16-x)) ?
The integral from-20 to 0 of 1/(sqrt(16-x)) is 4