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

The integral from 4 to 9 of (sqrt(x)+x) is 271/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 4 to 9 of (sqrt(x)+x)

\int_{4}^{9} (x + x) d x

\frac{271}{6}

Decimal

45.16666 \dots

Solution steps

\int_{4}^{9} x + x d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{4}^{9} x d x + \int_{4}^{9} x d x

\int_{4}^{9} x d x = \frac{38}{3}

\int_{4}^{9} x d x = \frac{65}{2}

= \frac{38}{3} + \frac{65}{2}

Simplify

\frac{38}{3} + \frac{65}{2} : \frac{271}{6}

= \frac{271}{6}

\int_{1}^{9} \frac{e ^{x}}{x} d x

\int_{a}^{x} \frac{1}{x} d x

\int_{- 2}^{2} (1 + x + x^{2} + x^{3}) d x

\int_{0}^{b} 3 e^{x} d x

\int_{4}^{8} 8 x - x^{2} d x

What is the integral from 4 to 9 of (sqrt(x)+x) ?
The integral from 4 to 9 of (sqrt(x)+x) is 271/6