What is the integral from 5 to 9 of (sqrt(9-y)+2)^2 ?

The integral from 5 to 9 of (sqrt(9-y)+2)^2 is 136/3

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integral from 5 to 9 of (sqrt(9-y)+2)^2

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

\frac{136}{3}

Decimal Notation

45.33333 \dots

Solution steps

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

9 - y + 2 = 2 (\frac{9 - y}{2} + 1)

= \int_{5}^{9} (2 (\frac{9 - y}{2} + 1))^{2} d y

Simplify

(2 (\frac{9 - y}{2} + 1))^{2} : - y + 4 - y + 9 + 13

= \int_{5}^{9} - y + 4 - y + 9 + 13 d y

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{5}^{9} y d y + \int_{5}^{9} 4 - y + 9 d y + \int_{5}^{9} 13 d y

\int_{5}^{9} y d y = 28

\int_{5}^{9} 4 - y + 9 d y = \frac{64}{3}

\int_{5}^{9} 13 d y = 52

= - 28 + \frac{64}{3} + 52

Simplify

- 28 + \frac{64}{3} + 52 : \frac{136}{3}

= \frac{136}{3}

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What is the integral from 5 to 9 of (sqrt(9-y)+2)^2 ?
The integral from 5 to 9 of (sqrt(9-y)+2)^2 is 136/3