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

The integral from 0 to 1 of (x^2+x+1) is 11/6

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 of (x^2+x+1)

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

\frac{11}{6}

Decimal

1.83333 \dots

Solution steps

\int_{0}^{1} x^{2} + x + 1 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_{0}^{1} x^{2} d x + \int_{0}^{1} x d x + \int_{0}^{1} 1 d x

\int_{0}^{1} x^{2} d x = \frac{1}{3}

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

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

= \frac{1}{3} + \frac{1}{2} + 1

Simplify

\frac{1}{3} + \frac{1}{2} + 1 : \frac{11}{6}

= \frac{11}{6}

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

\int_{1}^{5} 7 d x

\int_{0}^{t} cos (t - x) d x

\int_{0}^{l n (2)} e^{x} (1 + e^{x})^{3} d x

\int_{0}^{1} 5 x^{4} (5 - x^{5}) d x

What is the integral from 0 to 1 of (x^2+x+1) ?
The integral from 0 to 1 of (x^2+x+1) is 11/6