What is the integral from-1 to 0 of (3+sqrt(1-x^2)) ?

The integral from-1 to 0 of (3+sqrt(1-x^2)) is 3+pi/4

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

\int_{- 1}^{0} (3 + 1 - x^{2}) d x

3 + \frac{π}{4}

Decimal

3.78539 \dots

Solution steps

\int_{- 1}^{0} 3 + 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} 3 d x + \int_{- 1}^{0} 1 - x^{2} d x

\int_{- 1}^{0} 3 d x = 3

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

= 3 + \frac{π}{4}

\int_{0}^{4} e^{0.5 x} d x

\int_{1}^{4} \frac{1}{( x - 1 ) ^{3}} d x

\int_{0}^{\frac{π}{4}} 1 + sin^{2} (x) d x

\int_{- 1}^{3} \frac{3}{6 - 2 x} d x

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

What is the integral from-1 to 0 of (3+sqrt(1-x^2)) ?
The integral from-1 to 0 of (3+sqrt(1-x^2)) is 3+pi/4