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Setting up the Riemann sum for the definite integral

Consider the function f(x)=4−x2f(x) = 4 - x^2 on the interval [−2,2][-2, 2]. We want to express the area under the curve (above the xx-axis) using the limit definition of the definite integral, using nn subintervals of equal width with right endpoints as sample points.

Divide [−2,2][-2, 2] into nn subintervals of equal width, and let xi∗=xix_i^* = x_i be the right endpoint of the ii-th subinterval.

  1. Find Δx\Delta x in terms of nn.
  2. Find a formula for xix_i in terms of ii and nn.
  3. Write ∫−22(4−x2) dx\displaystyle\int_{-2}^{2} (4 - x^2)\, dx as lim⁡n→∞∑i=1nf(xi)Δx\displaystyle\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i) \Delta x, substituting your expressions for f(xi)f(x_i) and Δx\Delta x.

You do not need to evaluate the sum or the limit — just set up the expression.

Setting up the Riemann sum for the definite integral graph
Hint

The interval has total length 44, so Δx=4/n\Delta x = 4/n. The right endpoint of the ii-th subinterval starts at the left endpoint −2-2 and adds ii widths of Δx\Delta x. Substitute this xix_i into f(x)=4−x2f(x) = 4 - x^2 to get f(xi)f(x_i), then multiply by Δx\Delta x and sum.

Step-by-step solution

Since the interval [−2,2][-2,2] has length 44 and is divided into nn equal subintervals, the width of each subinterval is Δx=4n\Delta x = \dfrac{4}{n}. Starting from the left endpoint a=−2a = -2, the right endpoint of the ii-th subinterval is xi=−2+iΔx=−2+4inx_i = -2 + i\Delta x = -2 + \dfrac{4i}{n}.

Substituting into f(x)=4−x2f(x) = 4 - x^2 gives f(xi)=4−(−2+4in)2f(x_i) = 4 - \left(-2 + \dfrac{4i}{n}\right)^2. Therefore the definite integral is expressed as ∫−22(4−x2) dx=lim⁡n→∞∑i=1n[4−(−2+4in)2]⋅4n.\int_{-2}^{2} (4-x^2)\,dx = \lim_{n\to\infty} \sum_{i=1}^{n} \left[4 - \left(-2 + \frac{4i}{n}\right)^2\right] \cdot \frac{4}{n}. This is the required limit-of-Riemann-sums expression; no further simplification or evaluation is needed.

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