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

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

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

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

\frac{π}{2}

Solution steps

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

Expand

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

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

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

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

= 0 + \frac{π}{2}

Simplify

= \frac{π}{2}

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