en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Calculus >

integral from-1 to 1 of (1+x^2)

Solution

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

Solution

\frac{8}{3}

+1

Decimal

2.66666 \dots

Solution steps

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

\int_{- 1}^{1} 1 d x = 2

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

= 2 + \frac{2}{3}

Simplify

2 + \frac{2}{3} : \frac{8}{3}

= \frac{8}{3}

Graph

Popular Examples

integral from 0 to 1 of (-3x^2+4x+3)

integral from-1 to 2 of (x^3+1)

integral from-1 to 2 of 8-x^2-x

integral from-2 to 2 of sqrt(1-(x^2)/4)

integral from-2 to 1 of 2x^2

Frequently Asked Questions (FAQ)

What is the integral from-1 to 1 of (1+x^2) ?
The integral from-1 to 1 of (1+x^2) is 8/3