What is the integral from 0 to 2 of sqrt(4x+1) ?

The integral from 0 to 2 of sqrt(4x+1) is 13/3

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integral from 0 to 2 of sqrt(4x+1)

\int_{0}^{2} 4 x + 1 d x

\frac{13}{3}

Decimal Notation

4.33333 \dots

Solution steps

\int_{0}^{2} 4 x + 1 d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Compute the boundaries:

\frac{52}{3}

= \frac{1}{4} \cdot \frac{52}{3}

Simplify

= \frac{13}{3}

x \to 0 lim (\frac{ln ( 6 x + 1 )}{x})

\int \frac{r}{4 - r ^{2}} d r

\int (5 x - 1)^{3} d x

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

y^{^{''}} + 16 y = 2 sin (2 t), y (0) = 1, y^{^{'}} (0) = 0

What is the integral from 0 to 2 of sqrt(4x+1) ?
The integral from 0 to 2 of sqrt(4x+1) is 13/3