How do you evaluate the definite integral by the limit definition given #intsqrt(r^2-x^2)dx# from [-r,r]?

Question

Paisley Johnson · Accepted Answer

This can be integrated geometrically.

The integrand #f(x) = sqrt(r^2-x^2)# has graph #y = sqrt(r^2-x^2)# which is the upper semicircle for the circle #x^2+y^2 = r^2#. The integral from #-r# to #r# is the area of the semicircle. #int_-r^r sqrt(r^2-x^2) dx = 1/2 pir^2#

Axel Alvarez · Answer

undefined To evaluate the definite integral ∫sqrt(r^2-x^2)dx from -r to r using the limit definition, follow these steps:

1. Begin with the given integral: ∫sqrt(r^2-x^2)dx.
2. To apply the limit definition, divide the interval [-r, r] into subintervals of equal width. Let Δx represent the width of each subinterval.
3. Choose a representative point xi in each subinterval.
4. Form the Riemann sum: R = Σsqrt(r^2-xi^2)Δx, where the sum is taken over all subintervals.
5. Take the limit of the Riemann sum as the width of the subintervals approaches zero: lim(Δx→0) R.
6. This limit will converge to the definite integral ∫sqrt(r^2-x^2)dx over the interval [-r, r].

By applying the limit definition of the definite integral, you'll arrive at the desired result without introducing additional complexities.