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

The integral from 0 to 1 of sqrt(3x+4) is (14sqrt(7)-16)/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 0 to 1 of sqrt(3x+4)

\int_{0}^{1} 3 x + 4 d x

\frac{14 7 - 16}{9}

Decimal

2.33783 \dots

Solution steps

\int_{0}^{1} 3 x + 4 d x

Apply u-substitution

= \int_{4}^{7} \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{4}^{7} u d u

Apply the Power Rule

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{4}^{7}

Compute the boundaries:

\frac{14 7}{3} - \frac{16}{3}

= \frac{1}{3} (\frac{14 7}{3} - \frac{16}{3})

Simplify

= \frac{14 7 - 16}{9}

What is the integral from 0 to 1 of sqrt(3x+4) ?
The integral from 0 to 1 of sqrt(3x+4) is (14sqrt(7)-16)/9