What is the integral from 0 to 5 of (2sqrt(x)+3)^2 ?

The integral from 0 to 5 of (2sqrt(x)+3)^2 is 95+40sqrt(5)

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integral from 0 to 5 of (2sqrt(x)+3)^2

\int_{0}^{5} (2 x + 3)^{2} d x

95 + 405

Decimal Notation

184.44271 \dots

Solution steps

\int_{0}^{5} (2 x + 3)^{2} d x

Expand

(2 x + 3)^{2} : 4 x + 12 x + 9

= \int_{0}^{5} 4 x + 12 x + 9 d x

Apply the Sum Rule:

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

= \int_{0}^{5} 4 x d x + \int_{0}^{5} 12 x d x + \int_{0}^{5} 9 d x

\int_{0}^{5} 4 x d x = 50

\int_{0}^{5} 12 x d x = 405

\int_{0}^{5} 9 d x = 45

= 50 + 405 + 45

Simplify

= 95 + 405

\int_{0}^{9} y + 14 d y

\int_{0}^{17} 85 - 4 x d x

\int_{0}^{5} r^{2} d r

\int_{- 2}^{3} x + 4 - (x^{2} - 2) d x

\int_{0}^{3} x^{3} - 4 x^{2} + 4 x d x

What is the integral from 0 to 5 of (2sqrt(x)+3)^2 ?
The integral from 0 to 5 of (2sqrt(x)+3)^2 is 95+40sqrt(5)