What is the integral from 0 to 2 of 9sqrt(4x+1) ?

The integral from 0 to 2 of 9sqrt(4x+1) is 39

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 2 of 9sqrt(4x+1)

\int_{0}^{2} 9 4 x + 1 d x

39

Solution steps

\int_{0}^{2} 9 4 x + 1 d x

Take the constant out:

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

= 9 \cdot \int_{0}^{2} 4 x + 1 d x

Apply u-substitution

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

= 9 \cdot \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

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

Apply the Power Rule

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

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

Simplify

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

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

Compute the boundaries:

\frac{52}{3}

= \frac{9}{4} \cdot \frac{52}{3}

Simplify

= 39

\int_{0}^{2} \frac{1}{x ^{3} - x} d x

\int_{6}^{22} 2 f (\frac{x}{2}) d x

\int_{2}^{0} x d x

\int_{4}^{6} \frac{1}{x ^{2} - 4} d x

\int_{0}^{7} π x d x

What is the integral from 0 to 2 of 9sqrt(4x+1) ?
The integral from 0 to 2 of 9sqrt(4x+1) is 39