What is the integral from 0 to 4 of x/(sqrt(16-x^2)) ?

The integral from 0 to 4 of x/(sqrt(16-x^2)) is 4

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integral from 0 to 4 of x/(sqrt(16-x^2))

\int_{0}^{4} \frac{x}{16 - x ^{2}} d x

4

Solution steps

\int_{0}^{4} \frac{x}{16 - x ^{2}} d x

Apply u-substitution

: \int_{16}^{0} - \frac{1}{2 u} d u

= \int_{16}^{0} - \frac{1}{2 u} d u

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

= - \int_{0}^{16} - \frac{1}{2 u} d u

Take the constant out:

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

= - (- \frac{1}{2} \cdot \int_{0}^{16} \frac{1}{u} d u)

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

8

= \frac{1}{2} \cdot 8

Simplify

= 4

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What is the integral from 0 to 4 of x/(sqrt(16-x^2)) ?
The integral from 0 to 4 of x/(sqrt(16-x^2)) is 4