What is the integral from-1 to 0 of (4+sqrt(1-x^2)) ?

The integral from-1 to 0 of (4+sqrt(1-x^2)) is 4+pi/4

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

\int_{- 1}^{0} (4 + 1 - x^{2}) d x

4 + \frac{π}{4}

Decimal

4.78539 \dots

Solution steps

\int_{- 1}^{0} 4 + 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}^{0} 4 d x + \int_{- 1}^{0} 1 - x^{2} d x

\int_{- 1}^{0} 4 d x = 4

\int_{- 1}^{0} 1 - x^{2} d x = \frac{π}{4}

= 4 + \frac{π}{4}

\int_{0}^{4} π (x - 1) d x

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What is the integral from-1 to 0 of (4+sqrt(1-x^2)) ?
The integral from-1 to 0 of (4+sqrt(1-x^2)) is 4+pi/4