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

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

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

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

\frac{1}{2} + \frac{π}{4}

Decimal Notation

1.28539 \dots

Solution steps

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

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

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

= \frac{1}{2} + \frac{π}{4}

\int 1 0^{x} (2^{- x} + 5^{- x}) d x

\frac{d}{d x} (A x^{2} e^{2 x})

\int e^{x} - x^{\frac{2}{5}} d x

\frac{3}{2 s + 1} \cdot \frac{1}{s}

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

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