Definition of the Definite Integral — Question 10

Question 10 - Solution

Partition the interval $[0,\pi]$ into $n$ equal subintervals. Each subinterval has width $$\Delta x=\frac{\pi}{n}.$$

The midpoint of the $i$th subinterval is $$x_i^*=\left(i-\frac12\right)\Delta x =\left(i-\frac12\right)\frac{\pi}{n}.$$

Evaluate the function at each midpoint: $$f(x_i^*)=\sin\!\left(\left(i-\frac12\right)\frac{\pi}{n}\right).$$

The Riemann sum is $$\sum_{i=1}^{n} f(x_i^*)\,\Delta x = \frac{\pi}{n} \sum_{i=1}^{n} \sin\!\left(\left(i-\frac12\right)\frac{\pi}{n}\right).$$

Thus, $$\int_{0}^{\pi}\sin x\,dx = \lim_{n\to\infty} \frac{\pi}{n} \sum_{i=1}^{n} \sin\!\left(\left(i-\frac12\right)\frac{\pi}{n}\right).$$

Evaluating the definite integral directly, $$\int_{0}^{\pi}\sin x\,dx = \bigl[-\cos x\bigr]_{0}^{\pi} = (-\cos\pi)-(-\cos 0) = 2.$$

$$\boxed{2}$$