Calculus

∫ [ x - sqrt ( 25 - x^2)] dx from x = -5 to x = 5
= ∫ x dx from x = - 5 to x = 5 minus ∫ sqrt(25 - x^2)dx from x = - 5 to x = 5
(i) The first part can be interpreted as the area between line y = x and the x - axis from x = - 5 to x = 5.
Since the area from x = - 5 to x = 0 is the same as the area from x = 0 to x = 5 but one is -ve and the other is +ve. So net value = 0.
(ii) The second part can be interpreted as the area between y = sqrt(25 - x^2) from x = - 5 to x = 5.
y = sqrt(25 - x^2)
y^2 = 25 - x^2
x^2 + y^2 = 25 which is a circle with center (0,0) and radius 5.
So the second integral can be interpreted as the area of the semi-circle = 25π/2. ( + sqrt(25 - x^2) is the upper semi-circle and - sqrt(25 - x^2) is the lower semi-circle.)
In conclusion, the value of the integral = 0 - 25π/2 = - 25π/2.