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Setting up the area as a limit of midpoint Riemann sums

Consider the region under the curve f(x)=sin⁡(x)f(x) = \sin(x) on the interval [0,π][0, \pi].

The area under f(x)=sin⁡(x)f(x) = \sin(x) on [0,π][0, \pi] is defined as a definite integral, which can be expressed as a limit of Riemann sums using midpoints of nn equal subintervals.

Setting up the area as a limit of midpoint Riemann sums graph

Let nn be the number of subintervals of equal width, and let xi∗x_i^* denote the midpoint of the ii-th subinterval. Write Δx\Delta x (the width of each subinterval) explicitly in terms of nn, write xi∗x_i^* explicitly in terms of ii and nn, and then write the full expression ∫0πsin⁡(x) dx=lim⁡n→∞∑i=1nf(xi∗) Δx\int_0^\pi \sin(x)\, dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*)\, \Delta x as an explicit limit of a sum, substituting in f(xi∗)f(x_i^*), Δx\Delta x, and xi∗x_i^*.

Hint

Divide [0,π][0,\pi] into nn equal pieces to find Δx\Delta x. The right endpoint of the ii-th subinterval is 0+iΔx0 + i\Delta x; the midpoint is halfway between that and the left endpoint, i.e. subtract half of Δx\Delta x.

Step-by-step solution

Since the interval [0,π][0,\pi] has length π\pi and is divided into nn equal subintervals, the width of each subinterval is Δx=πn\Delta x = \dfrac{\pi}{n}. The ii-th subinterval is [xi−1,xi][x_{i-1}, x_i] where xi=iΔx=iπnx_i = i\Delta x = \dfrac{i\pi}{n}, so its left endpoint is xi−1=(i−1)πnx_{i-1} = \dfrac{(i-1)\pi}{n}.

The midpoint of the ii-th subinterval is the average of its endpoints: xi∗=12((i−1)πn+iπn)=(2i−1)π2nx_i^* = \dfrac{1}{2}\left(\dfrac{(i-1)\pi}{n} + \dfrac{i\pi}{n}\right) = \dfrac{(2i-1)\pi}{2n}. Substituting f(xi∗)=sin⁡(xi∗)f(x_i^*) = \sin(x_i^*) and Δx=π/n\Delta x = \pi/n into the Riemann sum formula gives ∫0πsin⁡(x) dx=lim⁡n→∞∑i=1nsin⁡ ⁣((2i−1)π2n)⋅πn.\int_0^\pi \sin(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n \sin\!\left(\frac{(2i-1)\pi}{2n}\right)\cdot \frac{\pi}{n}.

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