Interpret the definite integrals as sums and differences of areas and then use simple geometric formulas to evaluate the integrals.

In each case sketch the graph of the function. Keep in mind that the area bounded below the X-axis and above the graph of y=f(x)contributes as a negative number.

Integral from (0)^(1) root(1-x^2) dx

(0 is suppose to be on the bottom and 1 is suppose to be on the top for the integral part)

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We have `y=+-sqrt(1-x^2)` `implies x^2 + y^2 = 1`

This is a circle with radius ` `1 centred on the origin. Since we are evaluating the integral for `x` in the interval [0,1], this is one half of the circle. The area is

`(pir^2)/2 = pi/2`

`therefore`

`int_0^1sqrt(1-x^2)dx = pi/2`

**Area/integral = pi/2**

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