What is the integral from 0 to 2 of y/(sqrt(9-y^2)) ?

The integral from 0 to 2 of y/(sqrt(9-y^2)) is 3-sqrt(5)

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integral from 0 to 2 of y/(sqrt(9-y^2))

\int_{0}^{2} \frac{y}{9 - y ^{2}} d y

3 - 5

Decimal Notation

0.76393 \dots

Solution steps

\int_{0}^{2} \frac{y}{9 - y ^{2}} d y

Apply u-substitution

: \int_{9}^{5} - \frac{1}{2 u} d u

= \int_{9}^{5} - \frac{1}{2 u} d u

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

= - \int_{5}^{9} - \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_{5}^{9} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{5}^{9}

= - (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{5}^{9})

Simplify

- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{5}^{9}) : \frac{1}{2} [2 u]_{5}^{9}

= \frac{1}{2} [2 u]_{5}^{9}

Compute the boundaries:

6 - 25

= \frac{1}{2} (6 - 25)

Simplify

= 3 - 5

\int \frac{( 5 x + 3 )}{x ^{3} - 2 x ^{2} - 3 x} d x

y^{^{'}} = x^{2} \cdot y

x \to 2 + lim (5)

\int \frac{8 x ^{2} - 13 x + 2}{x ^{3} - 3 x ^{2} + 2 x} d x

\int \frac{sin ^{7} ( 8 x )}{cos ^{4} ( 8 x )} d x

What is the integral from 0 to 2 of y/(sqrt(9-y^2)) ?
The integral from 0 to 2 of y/(sqrt(9-y^2)) is 3-sqrt(5)