Let 2 C = sin dx. As f(x) = sin(1/x) is continuous except at 0 and bounded, f is integrable on [-1,1]. Since the integrand is positive (never negative), C > 0. Put u = 1/x and make the change of varibles to u so that f(x) = sin(u)² and dx = -u-2du. When x = u = -1. Therefore, 1, и %3 = 1 and when x = -1, 2 sin(u)*(u-?)du ´sin(u) C = du < 0. Where does this argument go wrong?

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Riemann Sum

Riemann Sums is a special type of approximation of the area under a curve by dividing it into multiple simple shapes like rectangles or trapezoids and is used in integrals when finite sums are involved. Figuring out the area of a curve is complex hence this method makes it simple. Usually, we take the help of different integration methods for this purpose. This is one of the major parts of integral calculus. 

Riemann Integral

Bernhard Riemann's integral was the first systematic description of the integral of a function on an interval in the branch of mathematics known as real analysis.