What is the integral from 0 to 6 of sqrt(2x+1) ?

The integral from 0 to 6 of sqrt(2x+1) is (13sqrt(13)-1)/3

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integral from 0 to 6 of sqrt(2x+1)

\int_{0}^{6} 2 x + 1 d x

\frac{13 13 - 1}{3}

Decimal Notation

15.29072 \dots

Solution steps

\int_{0}^{6} 2 x + 1 d x

Apply u-substitution

: \int_{1}^{13} \frac{u}{2} d u

= \int_{1}^{13} \frac{u}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{1}^{13} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{13}

Compute the boundaries:

\frac{26 13}{3} - \frac{2}{3}

= \frac{1}{2} (\frac{26 13}{3} - \frac{2}{3})

Simplify

= \frac{13 13 - 1}{3}

x \to 0 lim (sin (x) \cdot \frac{1}{x})

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

x \to 8 lim (\frac{( x + 6 )}{x - 8})

\int \frac{x ^{2}}{( 5 + x ^{3} ) ^{2}} d x

f (x) = - 2 cos (x)

What is the integral from 0 to 6 of sqrt(2x+1) ?
The integral from 0 to 6 of sqrt(2x+1) is (13sqrt(13)-1)/3