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

The integral from 0 to x of sqrt(1+9/4 t) is 4/9 (((4+9x)^{3/2})/(12)-2/3)

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

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

\frac{4}{9} (\frac{( 4 + 9 x ) ^{\frac{3}{2}}}{12} - \frac{2}{3})

Solution steps

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

Apply u-substitution

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

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

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{1 + \frac{9 x}{4}}

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

Compute the boundaries:

\frac{( 4 + 9 x ) ^{\frac{3}{2}}}{12} - \frac{2}{3}

= \frac{4}{9} (\frac{( 4 + 9 x ) ^{\frac{3}{2}}}{12} - \frac{2}{3})

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What is the integral from 0 to x of sqrt(1+9/4 t) ?
The integral from 0 to x of sqrt(1+9/4 t) is 4/9 (((4+9x)^{3/2})/(12)-2/3)