What is the integral from 0 to x of sqrt(1+81/4 t) ?

The integral from 0 to x of sqrt(1+81/4 t) is 4/81 (((4+81x)^{3/2})/(12)-2/3)

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integral from 0 to x of sqrt(1+81/4 t)

\int_{0}^{x} 1 + \frac{81}{4} t d t

\frac{4}{81} (\frac{( 4 + 81 x ) ^{\frac{3}{2}}}{12} - \frac{2}{3})

Solution steps

\int_{0}^{x} 1 + \frac{81}{4} t d t

Apply u-substitution

: \int_{1}^{1 + \frac{81 x}{4}} \frac{4 u}{81} d u

= \int_{1}^{1 + \frac{81 x}{4}} \frac{4 u}{81} d u

Take the constant out:

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

= \frac{4}{81} \cdot \int_{1}^{1 + \frac{81 x}{4}} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{1 + \frac{81 x}{4}}

= \frac{4}{81} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{1 + \frac{81 x}{4}}

Compute the boundaries:

\frac{( 4 + 81 x ) ^{\frac{3}{2}}}{12} - \frac{2}{3}

= \frac{4}{81} (\frac{( 4 + 81 x ) ^{\frac{3}{2}}}{12} - \frac{2}{3})

\int (x)^{3} d x

f (x) = \frac{1}{x ^{2}}, at x = 1

\frac{\partial}{\partial x} (e^{- x y} cos (x y))

\frac{\partial}{\partial x} (5 e^{x^{2} - 4 y})

\frac{d}{d x} (\frac{x ^{2} + 4 x - 5}{ln ( x )})

What is the integral from 0 to x of sqrt(1+81/4 t) ?
The integral from 0 to x of sqrt(1+81/4 t) is 4/81 (((4+81x)^{3/2})/(12)-2/3)