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

The integral from 0 to 16 of 1/(sqrt(16-x)) is 8

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

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

8

Solution steps

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

Apply u-substitution

: \int_{16}^{0} - \frac{1}{u} d u

= \int_{16}^{0} - \frac{1}{u} d u

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

= - \int_{0}^{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_{0}^{16} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{0}^{16}

= - (- [2 u^{\frac{1}{2}}]_{0}^{16})

Simplify

- (- [2 u^{\frac{1}{2}}]_{0}^{16}) : [2 u]_{0}^{16}

= [2 u]_{0}^{16}

Compute the boundaries:

8

= 8

f (x) = (- 3 x^{2} + 3) e^{2 x}, at x = 2

\int \frac{1}{3 + x ^{2}} d x

\int (\frac{x ^{2}}{3} + 7 x) d x

\int_{- 1}^{1} 2 ∣ x ∣ d x

5 + \frac{6}{x} + \frac{6}{x ^{2}}

What is the integral from 0 to 16 of 1/(sqrt(16-x)) ?
The integral from 0 to 16 of 1/(sqrt(16-x)) is 8