What is the integral from 1 to 3 of sqrt(x+6) ?

The integral from 1 to 3 of sqrt(x+6) is 18-(14sqrt(7))/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 1 to 3 of sqrt(x+6)

\int_{1}^{3} x + 6 d x

18 - \frac{14 7}{3}

Decimal

5.65316 \dots

Solution steps

\int_{1}^{3} x + 6 d x

Apply u-substitution

: \int_{7}^{9} u d u

= \int_{7}^{9} u d u

Apply the Power Rule

: [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{9}

= [\frac{2}{3} u^{\frac{3}{2}}]_{7}^{9}

Compute the boundaries:

18 - \frac{14 7}{3}

= 18 - \frac{14 7}{3}

\int_{0}^{x} sin (8 u) d u

\int_{0}^{x^{2} + y^{2}} 1 d z

\int_{0}^{3} \frac{6}{7 x + 5} d x

\int_{4}^{12} (4 - x) (x - 12) d x

\int_{0}^{\frac{π}{3}} (sec (x))^{2} d x

What is the integral from 1 to 3 of sqrt(x+6) ?
The integral from 1 to 3 of sqrt(x+6) is 18-(14sqrt(7))/3