What is the integral from 0 to 1 of (x+6sqrt(1-x^2)) ?

The integral from 0 to 1 of (x+6sqrt(1-x^2)) is 1/2+(3pi)/2

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integral from 0 to 1 of (x+6sqrt(1-x^2))

\int_{0}^{1} (x + 6 1 - x^{2}) d x

\frac{1}{2} + \frac{3 π}{2}

Decimal

5.21238 \dots

Solution steps

\int_{0}^{1} x + 6 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_{0}^{1} x d x + \int_{0}^{1} 6 1 - x^{2} d x

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

\int_{0}^{1} 6 1 - x^{2} d x = \frac{3 π}{2}

= \frac{1}{2} + \frac{3 π}{2}

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What is the integral from 0 to 1 of (x+6sqrt(1-x^2)) ?
The integral from 0 to 1 of (x+6sqrt(1-x^2)) is 1/2+(3pi)/2