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

The integral from 0 to 9 of (sqrt(9-y)+2)^2 is 297/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

\int_{0}^{9} (9 - y + 2)^{2} d y

\frac{297}{2}

Decimal

148.5

Solution steps

\int_{0}^{9} (9 - y + 2)^{2} d y

Multiply by the conjugate of

9 - y + 2 : \frac{- y + 5}{9 - y - 2}

= \int_{0}^{9} (\frac{- y + 5}{9 - y - 2})^{2} d y

Find undefined (singularity) points:

y = 5

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{5} (\frac{- y + 5}{9 - y - 2})^{2} d y + \int_{5}^{9} (\frac{- y + 5}{9 - y - 2})^{2} d y

\int_{0}^{5} (\frac{- y + 5}{9 - y - 2})^{2} d y = \frac{619}{6}

\int_{5}^{9} (\frac{- y + 5}{9 - y - 2})^{2} d y = \frac{136}{3}

= \frac{619}{6} + \frac{136}{3}

Simplify

\frac{619}{6} + \frac{136}{3} : \frac{297}{2}

= \frac{297}{2}

9 x y^{^{'}} = 2 x e^{x} - 9 y + 6 x^{2}

\frac{2 s + 6}{s ^{2} + 4}

x^{4} + 5 y^{4} x^{5} + 4 y^{^{'}} = 0

\frac{1}{3 s - 2}

\frac{e ^{x}}{1 - x}

What is the integral from 0 to 9 of (sqrt(9-y)+2)^2 ?
The integral from 0 to 9 of (sqrt(9-y)+2)^2 is 297/2