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

The integral from 0 to 2 of y/(sqrt(25-y^2)) is 5-sqrt(21)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

5 - 21

Decimal Notation

0.41742 \dots

Solution steps

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

Apply u-substitution

: \int_{25}^{21} - \frac{1}{2 u} d u

= \int_{25}^{21} - \frac{1}{2 u} d u

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

= - \int_{21}^{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_{21}^{25} \frac{1}{u} d u)

Apply the Power Rule

: [2 u^{\frac{1}{2}}]_{21}^{25}

= - (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{21}^{25})

Simplify

- (- \frac{1}{2} [2 u^{\frac{1}{2}}]_{21}^{25}) : \frac{1}{2} [2 u]_{21}^{25}

= \frac{1}{2} [2 u]_{21}^{25}

Compute the boundaries:

10 - 221

= \frac{1}{2} (10 - 221)

Simplify

= 5 - 21

x \to - \infty lim (\frac{x ^{6} + cos ( x ^{10000} )}{x + 5})

f (x) = x^{11}

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

- \frac{x}{2 y}

\int 1 - cos (2 x) d x

What is the integral from 0 to 2 of y/(sqrt(25-y^2)) ?
The integral from 0 to 2 of y/(sqrt(25-y^2)) is 5-sqrt(21)