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

The integral from 0 to 5 of sqrt(1+9/4 x) is 335/27

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integral from 0 to 5 of sqrt(1+9/4 x)

\int_{0}^{5} 1 + \frac{9}{4} x d x

\frac{335}{27}

Decimal

12.40740 \dots

Solution steps

\int_{0}^{5} 1 + \frac{9}{4} x d x

Apply u-substitution

: \int_{1}^{\frac{49}{4}} \frac{4 u}{9} d u

= \int_{1}^{\frac{49}{4}} \frac{4 u}{9} d u

Take the constant out:

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

= \frac{4}{9} \cdot \int_{1}^{\frac{49}{4}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{49}{4}}

= \frac{4}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{\frac{49}{4}}

Compute the boundaries:

\frac{335}{12}

= \frac{4}{9} \cdot \frac{335}{12}

Simplify

= \frac{335}{27}

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What is the integral from 0 to 5 of sqrt(1+9/4 x) ?
The integral from 0 to 5 of sqrt(1+9/4 x) is 335/27