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

The integral from 0 to 1 of sqrt(4x+3) is (7sqrt(7)-3sqrt(3))/6

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(4x+3)

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

\frac{7 7 - 3 3}{6}

Decimal

2.22068 \dots

Solution steps

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

Apply u-substitution

= \int_{3}^{7} \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_{3}^{7} u d u

Apply the Power Rule

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

Compute the boundaries:

\frac{14 7}{3} - 23

= \frac{1}{4} (\frac{14 7}{3} - 23)

Simplify

= \frac{7 7 - 3 3}{6}

What is the integral from 0 to 1 of sqrt(4x+3) ?
The integral from 0 to 1 of sqrt(4x+3) is (7sqrt(7)-3sqrt(3))/6