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

The integral from-1 to 1 of (3+sqrt(1-x^2)) is 6+pi/2

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

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

6 + \frac{π}{2}

Solution steps

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

\int_{- 1}^{1} 3 d x = 6

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

= 6 + \frac{π}{2}

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

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

\int_{0}^{10} x^{3} d x

\int_{0}^{t} \frac{sin ( u )}{u} d u

\int_{\frac{π}{2}}^{π} x d x

What is the integral from-1 to 1 of (3+sqrt(1-x^2)) ?
The integral from-1 to 1 of (3+sqrt(1-x^2)) is 6+pi/2