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

The integral from-1 to 1 of (1+x)sqrt(1-x^2) is pi/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

\frac{π}{2}

Decimal

1.57079 \dots

Solution steps

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

Expand

(1 + x) 1 - x^{2} : 1 - x^{2} + x 1 - x^{2}

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

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

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

= \frac{π}{2} + 0

Simplify

= \frac{π}{2}

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