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Setting up the right-endpoint Riemann sum for area under tan(x)

Consider the function f(x)=tan⁡(x)f(x) = \tan(x) on the interval [0,π/4][0, \pi/4]. Since f(x)≥0f(x) \geq 0 on this interval, the definite integral ∫0π/4tan⁡(x) dx\int_0^{\pi/4} \tan(x)\,dx represents the area under the curve.

Suppose the interval [0,π/4][0, \pi/4] is divided into nn subintervals of equal width, and xi∗x_i^* denotes the right endpoint of the ii-th subinterval. Write down Δx\Delta x and xi∗x_i^* in terms of nn and ii, and use them to express the area under f(x)=tan⁡(x)f(x) = \tan(x) on [0,π/4][0,\pi/4] as a limit of a Riemann sum (do not evaluate the limit).

Hint

Divide the interval length π/4−0\pi/4 - 0 into nn equal pieces to get Δx\Delta x, then the right endpoint of the ii-th subinterval is the left endpoint 00 plus ii widths of Δx\Delta x. Substitute this xi∗x_i^* into f(x)=tan⁡(x)f(x) = \tan(x) and sum over i=1i=1 to nn, then take n→∞n \to \infty.

Step-by-step solution

Since the interval [0,π/4][0,\pi/4] has length π/4\pi/4, dividing it into nn equal subintervals gives width Δx=π/4−0n=π4n\Delta x = \dfrac{\pi/4 - 0}{n} = \dfrac{\pi}{4n}. The right endpoint of the ii-th subinterval, starting from a=0a=0, is xi∗=0+iΔx=iπ4nx_i^* = 0 + i\Delta x = \dfrac{i\pi}{4n}.

Substituting into the general definition ∫abf(x) dx=lim⁡n→∞∑i=1nf(xi∗)Δx\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x, we replace ff with tan⁡\tan and use the values of xi∗x_i^* and Δx\Delta x found above. This gives the area as lim⁡n→∞∑i=1ntan⁡ ⁣(iπ4n)⋅π4n.\lim_{n\to\infty}\sum_{i=1}^{n}\tan\!\left(\frac{i\pi}{4n}\right)\cdot\frac{\pi}{4n}. This limit expression, not a numerical evaluation, is the desired answer.

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