`int_-5^5(x - sqrt(25 - x^2))dx` Evaluate the integral by interpreting it in terms of areas.
I_1 can be be interpreted as area of two triangles;one above the x-axis and the other below axis.Since they are on the opposite sides of x-axis so the net resultant area will be zero.(see attached graph)
I_2 can be interpreted as the top half of a circle with centre the origin and radius 5 and above the x-axis from x=-5 to x=5.
Area of semi circle=`pi(5^2)/2=(25pi)/2`