What is the integral from 0 to 9 of sqrt(4x+9) ?

The integral from 0 to 9 of sqrt(4x+9) is (9(5sqrt(5)-1))/2

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integral from 0 to 9 of sqrt(4x+9)

\int_{0}^{9} 4 x + 9 d x

\frac{9 ( 5 5 - 1 )}{2}

Decimal Notation

45.81152 \dots

Solution steps

\int_{0}^{9} 4 x + 9 d x

Apply u-substitution

: \int_{9}^{45} \frac{u}{4} d u

= \int_{9}^{45} \frac{u}{4} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int_{9}^{45} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{45}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{45}

Compute the boundaries:

905 - 18

= \frac{1}{4} (905 - 18)

Simplify

= \frac{9 ( 5 5 - 1 )}{2}

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What is the integral from 0 to 9 of sqrt(4x+9) ?
The integral from 0 to 9 of sqrt(4x+9) is (9(5sqrt(5)-1))/2