What is the integral from 0 to 4 of y/(sqrt(25-y^2)) ?

The integral from 0 to 4 of y/(sqrt(25-y^2)) is 2

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integral from 0 to 4 of y/(sqrt(25-y^2))

\int_{0}^{4} \frac{y}{25 - y ^{2}} d y

2

Solution steps

\int_{0}^{4} \frac{y}{25 - y ^{2}} d y

Apply u-substitution

: \int_{25}^{9} - \frac{1}{2 u} d u

= \int_{25}^{9} - \frac{1}{2 u} d u

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

= - \int_{9}^{25} - \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_{9}^{25} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{9}^{25}

= - (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{25})

Simplify

- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{9}^{25}) : \frac{1}{2} [2 u]_{9}^{25}

= \frac{1}{2} [2 u]_{9}^{25}

Compute the boundaries:

4

= \frac{1}{2} \cdot 4

Simplify

= 2

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What is the integral from 0 to 4 of y/(sqrt(25-y^2)) ?
The integral from 0 to 4 of y/(sqrt(25-y^2)) is 2