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

The integral from 1 to 4 of sqrt(1+(9x)/4) is (80sqrt(10)-13sqrt(13))/(27)

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

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

\frac{80 10 - 13 13}{27}

Decimal Notation

7.63370 \dots

Solution steps

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

Apply u-substitution

: \int_{\frac{13}{4}}^{10} \frac{4 u}{9} d u

= \int_{\frac{13}{4}}^{10} \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_{\frac{13}{4}}^{10} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{13}{4}}^{10}

= \frac{4}{9} [\frac{2}{3} u^{\frac{3}{2}}]_{\frac{13}{4}}^{10}

Compute the boundaries:

\frac{20 10}{3} - \frac{13 13}{12}

= \frac{4}{9} (\frac{20 10}{3} - \frac{13 13}{12})

Simplify

= \frac{80 10 - 13 13}{27}

\frac{\partial}{\partial y} (5 x^{2} y + 5 x y^{2})

\frac{d}{d x} (x (x - 1)^{2})

y^{^{'}} = \frac{( 1 + y )}{x ^{2} + x}

\frac{1}{3 s}

\frac{\partial}{\partial x} (x^{2} y z^{2})

What is the integral from 1 to 4 of sqrt(1+(9x)/4) ?
The integral from 1 to 4 of sqrt(1+(9x)/4) is (80sqrt(10)-13sqrt(13))/(27)