Linear Functions and Triangular Area

From Uniform Sums to Area

A straight-line graph makes the relationship between a Riemann sum and geometric area especially visible. On each subinterval, a Riemann sum uses a rectangle whose height is a function value. As the subintervals become narrower, the sum approaches the integral. For an affine function, we can calculate that limit directly and compare it with the area of the triangle or trapezoid under the graph.

The calculation will use right endpoints on a uniform partition. The definite integral is defined through limits of tagged Riemann sums, and the earlier material on Riemann integrability establishes that continuous functions are integrable and that the limit does not depend on the tag choices. Since an affine function is continuous, a convenient choice of tags is enough to find its integral.

The Affine Function and Its Uniform Sum

The factor \(\Delta x\) is the width of each rectangle. It is essential: the function value gives the rectangle's height, while the width times the height gives its contribution to the sum.

For the right endpoints, the function values form an arithmetic sequence. We will use the identity \(1+2+\cdots+n=n(n+1)/2\). To see why it holds, write the sum forward and backward: \(S=1+2+\cdots+n\) and \(S=n+(n-1)+\cdots+1\). Adding corresponding terms gives \(2S=n(n+1)\), so \(S=n(n+1)/2\).

Deriving the Integral Formula

For \(f(x)=mx+c\), the value at the right endpoint \(x_i=a+i\Delta x\) is \(f(x_i)=f(a)+mi\Delta x\). Substitute these values into the right-endpoint sum, use the arithmetic-series identity, and then let the number of subintervals tend to infinity.

As \(n\) increases without bound, \(1/n\) tends to zero. Because the affine function is Riemann integrable, the limit of these sums is its definite integral. Also, \(f(b)=f(a)+m(b-a)\), so the limit can be written using the two endpoint values.

The expression on the right is the interval width multiplied by the average of the function's endpoint values. Geometrically, that is the area formula for a trapezoid whose parallel sides have lengths \(f(a)\) and \(f(b)\), provided the graph lies above the horizontal axis throughout the interval. If one endpoint value is zero and the other is positive, the trapezoid is a triangle.

Worked Examples

Why the Geometric Check Works

On \([a,b]\), an affine graph is a straight segment joining \((a,f(a))\) to \((b,f(b))\). If both endpoint heights are nonnegative, the region under the graph and above the horizontal axis is a trapezoid, possibly with one side of length zero. Its area is the width times the average of the two parallel-side lengths:

This is exactly the value obtained from the Riemann-sum limit. When one endpoint height is zero and the other is positive, the trapezoid has a zero-length side and is a triangle. The integral therefore agrees with the familiar triangular-area calculation in that case.

The endpoint formula is an integral formula, even when a geometric area interpretation does not apply directly. For example, if the graph lies below the horizontal axis, the integral is negative rather than the positive geometric area. The formula still holds because it was derived from the Riemann sums. This tutorial's triangle and trapezoid comparisons use graphs that stay above the axis.

A Common Mistake: Forgetting the Width

For \(f(x)=x\) on \([0,2]\), the right endpoints are \(x_i=2i/n\). A common error is to add the function values but forget to multiply each one by the subinterval width. That incorrect calculation gives

This expression grows without bound, but it is not a Riemann sum: it omits the rectangle width \(\Delta x=2/n\). The correct sum is

Its limit is \(2\). The endpoint formula confirms the answer: \(\frac{2}{2}(f(0)+f(2))=0+2=2\). Always include the subinterval width in every term of a Riemann sum.