What is the integral from 0 to 1/2 of 6/7 (2x^2+x) ?

The integral from 0 to 1/2 of 6/7 (2x^2+x) is 5/28

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1/2 of 6/7 (2x^2+x)

\int_{0}^{\frac{1}{2}} \frac{6}{7} (2 x^{2} + x) d x

\frac{5}{28}

Decimal Notation

0.17857 \dots

Solution steps

\int_{0}^{\frac{1}{2}} \frac{6}{7} (2 x^{2} + x) d x

Take the constant out:

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

= \frac{6}{7} \cdot \int_{0}^{\frac{1}{2}} 2 x^{2} + 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

= \frac{6}{7} (\int_{0}^{\frac{1}{2}} 2 x^{2} d x + \int_{0}^{\frac{1}{2}} x d x)

\int_{0}^{\frac{1}{2}} 2 x^{2} d x = \frac{1}{12}

\int_{0}^{\frac{1}{2}} x d x = \frac{1}{8}

= \frac{6}{7} (\frac{1}{12} + \frac{1}{8})

Simplify

\frac{6}{7} (\frac{1}{12} + \frac{1}{8}) : \frac{5}{28}

= \frac{5}{28}

\int_{- 1}^{1} 2 e^{i αx} d x

\int_{0}^{x^{2}} t^{3} d t

\int_{0}^{\frac{π}{2}} sin (x^{4}) d x

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

\int_{0}^{π} sin (x) cos (4 x) d x

What is the integral from 0 to 1/2 of 6/7 (2x^2+x) ?
The integral from 0 to 1/2 of 6/7 (2x^2+x) is 5/28