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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{13}{3}

Decimal

4.33333 \dots

Solution steps

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

Apply u-substitution

= \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{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

t an g e n t \frac{3 x ^{2}}{( 3 - 2 x ) ^{4}}

\int (2 t e^{t}) d t

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