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

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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}

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