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

The integral from-1 to 1 of 1/6 (x+3) is 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 1 of 1/6 (x+3)

\int_{- 1}^{1} \frac{1}{6} (x + 3) d x

1

Solution steps

\int_{- 1}^{1} \frac{1}{6} (x + 3) d x

Take the constant out:

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

= \frac{1}{6} \cdot \int_{- 1}^{1} x + 3 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{1}{6} (\int_{- 1}^{1} x d x + \int_{- 1}^{1} 3 d x)

\int_{- 1}^{1} x d x = 0

\int_{- 1}^{1} 3 d x = 6

= \frac{1}{6} (0 + 6)

Simplify

\frac{1}{6} (0 + 6) : 1

= 1

\int_{0}^{5} 5 e^{x} d x

\int_{1}^{2} (x^{2} - 5 x) d x

\int_{0}^{a} 6 e^{x} d x

\int_{0}^{\frac{π}{3}} cos (x) - \frac{1}{2} d x

\int_{- 1}^{3} 4 - (x - 1)^{2} d x

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