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Setting up the midpoint Riemann sum as a limit

Consider the function f(x)=x4f(x) = x^4 on the interval [−1,1][-1, 1]. We want to express the area between the curve and the xx-axis as a limit of midpoint Riemann sums.

Suppose [−1,1][-1,1] is divided into nn subintervals of equal width, and xi∗x_i^* denotes the midpoint of the ii-th subinterval. Write the width Δx\Delta x in terms of nn, and express the exact area under f(x)=x4f(x) = x^4 on [−1,1][-1,1] as ∫−11x4 dx=lim⁡n→∞∑i=1nf(xi∗) Δx\int_{-1}^{1} x^4\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\,\Delta x by giving the explicit value of Δx\Delta x and the general form of f(xi∗)f(x_i^*) in terms of ii and nn (using xi∗=−1+(i−12)Δxx_i^* = -1 + \left(i - \tfrac12\right)\Delta x).

Hint

First find Δx=(b−a)/n\Delta x = (b-a)/n for a=−1a=-1, b=1b=1. Then substitute the midpoint formula xi∗=−1+(i−12)Δxx_i^* = -1 + (i-\tfrac12)\Delta x into f(x)=x4f(x) = x^4, and plug both Δx\Delta x and f(xi∗)f(x_i^*) into the Riemann sum expression — you do not need to evaluate the sum or the limit.

Step-by-step solution

The interval [−1,1][-1,1] has length b−a=2b - a = 2, so dividing into nn equal subintervals gives Δx=2n\Delta x = \dfrac{2}{n}. The midpoint of the ii-th subinterval is xi∗=−1+(i−12)Δx=−1+2i−1nx_i^* = -1 + \left(i - \tfrac12\right)\Delta x = -1 + \dfrac{2i-1}{n}.

Substituting into f(x)=x4f(x) = x^4 gives f(xi∗)=(−1+2i−1n)4f(x_i^*) = \left(-1 + \dfrac{2i-1}{n}\right)^4. Therefore the definite integral is expressed as ∫−11x4 dx=lim⁡n→∞∑i=1n(−1+2i−1n)4⋅2n.\int_{-1}^{1} x^4\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} \left(-1 + \frac{2i-1}{n}\right)^4 \cdot \frac{2}{n}. This is the required limit-of-Riemann-sums expression; no evaluation is needed.

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