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Setting up the Riemann sum for a constant function

Consider the constant function f(x)=5f(x) = 5 on the interval [2,6][2, 6].

Divide [2,6][2,6] into nn subintervals of equal width to form a Riemann sum for f(x)=5f(x) = 5. Write Δx\Delta x in terms of nn, and write f(xi∗)f(x_i^*) for an arbitrary sample point xi∗x_i^* in the ii-th subinterval. Then write the Riemann sum ∑i=1nf(xi∗) Δx\sum_{i=1}^{n} f(x_i^*)\,\Delta x as an explicit expression in terms of nn only (no summation notation), and take the limit as n→∞n \to \infty to find ∫265 dx\displaystyle\int_2^6 5\,dx. Finally, confirm your limit matches the area of the corresponding geometric shape (rectangle) with base [2,6][2,6] and height 55.

What is the value of ∫265 dx\displaystyle\int_2^6 5\,dx, and does it match the rectangle's area?

Hint

Since ff is constant, f(xi∗)=5f(x_i^*) = 5 regardless of where the sample point is chosen, so the sum simplifies dramatically before you even take the limit.

Step-by-step solution

The interval [2,6][2,6] has length 44, so Δx=4n\Delta x = \frac{4}{n}. Since f(x)=5f(x) = 5 for every xx, we have f(xi∗)=5f(x_i^*) = 5 for every sample point, regardless of choice. So the Riemann sum is ∑i=1nf(xi∗) Δx=∑i=1n5⋅4n=n⋅(5⋅4n)=20\sum_{i=1}^n f(x_i^*)\,\Delta x = \sum_{i=1}^n 5 \cdot \frac{4}{n} = n \cdot \left(5 \cdot \frac{4}{n}\right) = 20.

Taking the limit as n→∞n \to \infty, we get lim⁡n→∞20=20\lim_{n\to\infty} 20 = 20, so ∫265 dx=20\int_2^6 5\,dx = 20. Geometrically, the region under f(x)=5f(x)=5 over [2,6][2,6] is a rectangle with base 6−2=46-2 = 4 and height 55, giving area 4×5=204 \times 5 = 20. This matches the Riemann sum limit exactly, confirming the result.

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