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

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

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

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

2 + \frac{π}{4}

Decimal

2.78539 \dots

Solution steps

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

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

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

= 2 + \frac{π}{4}

\frac{d}{d x} ((1 + x)^{\frac{1}{x}})

\int sin (t^{2}) d t

f (x) = 4 x^{2} - 5 x, at x = 3

f (x) = - 4 x

\frac{\partial}{\partial y} (\frac{t}{2 y ^{4}})

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