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Setting up the right-endpoint Riemann sum limit for a definite integral

Consider the function f(x)=2xf(x) = 2^x on the interval [0,3][0, 3].

Divide [0,3][0,3] into nn subintervals of equal width and let xix_i denote the right endpoint of the ii-th subinterval. Express the area under f(x)=2xf(x) = 2^x on [0,3][0,3] as a limit of right-endpoint Riemann sums. That is, write ∫032x dx\displaystyle\int_0^3 2^x\,dx explicitly in the form

lim⁡n→∞∑i=1nf(xi) Δx\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\,\Delta x

by giving explicit formulas for Δx\Delta x, xix_i, and f(xi)f(x_i) in terms of nn and ii, and writing the full limit expression.

Hint

First find Δx=b−an\Delta x = \frac{b-a}{n} for this interval, then write xi=a+i Δxx_i = a + i\,\Delta x as the right endpoint of the ii-th subinterval. Substitute this xix_i into f(x)=2xf(x) = 2^x to get f(xi)f(x_i), then assemble the full sum inside the limit.

Step-by-step solution

Since [0,3][0,3] is divided into nn equal subintervals, the width of each subinterval is Δx=3−0n=3n\Delta x = \frac{3-0}{n} = \frac{3}{n}. The right endpoint of the ii-th subinterval starting at a=0a=0 is xi=0+i Δx=3inx_i = 0 + i\,\Delta x = \frac{3i}{n}.

Substituting this right endpoint into f(x)=2xf(x) = 2^x gives f(xi)=23i/nf(x_i) = 2^{3i/n}. Putting everything together, the definite integral is ∫032x dx=lim⁡n→∞∑i=1n23i/n⋅3n.\int_0^3 2^x\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} 2^{3i/n} \cdot \frac{3}{n}. This limit expresses the exact area under the curve as the limit of the sum of areas of nn approximating rectangles, each of height 23i/n2^{3i/n} and width 3n\frac{3}{n}.

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