Reverse or collapse integration bounds
A constant height, two directions
The line has the constant height 3. The integral from the start to the end is the signed area of the teal region between the line and the axis.
With the start at 1 and the end at 5, the region is 4 wide and 3 tall, so the integral is \(3\cdot4 = 12\).
Put the start at 5 and the end at 1. The arrow along the axis now runs from right to left, so the width is \(1 - 5 = -4\). The region turns red because it counts as negative, and the integral is \(3\cdot(-4) = -12\).
Put the start and the end at the same place. The region has no width, so the integral is 0.
Strips with a signed width
A Riemann sum adds the areas of thin strips. Each strip has the height \(f(x_i)\) of the curve at its start and the width \(\Delta x\).
The width is the end minus the start, divided by the number of strips \(n\). From \(a\) to \(b\) it is \(\Delta x = \frac{b - a}{n}\). The curve lies above the axis and the width is positive, so each strip counts as positive and is teal.
Press Swap the ends. The path now runs from \(b\) to \(a\). Each strip keeps its height, but its width becomes \(\frac{a - b}{n}\), which is negative, so every strip turns red.
Every term of the sum changes sign, so the sum changes sign, and so does its limit, the integral.
Drag \(b\) toward \(a\). The strips become thinner, and when \(b\) reaches \(a\) the width is 0, so every term is 0 and the integral is 0.
Notation
The width of a strip is the end minus the start, divided by the number of strips \(n\). It is negative when the path runs from right to left.
\[\cg{\Delta x = \dfrac{b - a}{n}}\]If you swap the bounds, each width changes sign, so each term of the sum changes sign, and so does the integral.
\[\cr{\int_b^a f(x)\,dx} = -\int_a^b f(x)\,dx\]If the bounds are equal, the width is 0, so every term is 0.
\[\int_a^a f(x)\,dx = 0\]This needs \(f\) to be defined at \(a\).
Both rules hold for any function that has an integral on the interval, whether its curve lies above the axis or below it.
Example 1: bounds in the wrong order
Evaluate \(\displaystyle\int_4^1 2x\,dx\).
Write the bounds in order, with the smaller number at the bottom. The region under the line from 1 to 4 is a trapezoid with parallel sides \(f(1) = 2\) and \(f(4) = 8\).
The trapezoid is 3 wide, and its area is the average of the two parallel sides times the width.
\[\int_1^4 2x\,dx = \tfrac{2 + 8}{2}\cdot 3 = \cb{15}\]The given integral runs from 4 to 1, from right to left, so the red region counts as negative and the integral changes sign.
\[\int_4^1 2x\,dx = -\int_1^4 2x\,dx = \cr{-15}\]
Example 2: a region below the axis
Evaluate \(\displaystyle\int_3^1 (-2x)\,dx\).
Write the bounds in order. The line lies below the axis from 1 to 3, so the region is red and counts as negative.
\[\int_1^3 (-2x)\,dx = \tfrac{-2 + (-6)}{2}\cdot 2 = \cr{-8}\]A tempting wrong move: The region lies below the axis, so \(\displaystyle\int_3^1 (-2x)\,dx = -8\).
The bounds run from 3 to 1, from right to left, so the width is negative. A negative height times a negative width is positive, so the region turns teal and cannot count as −8.
Swap the bounds and change the sign of the result.
\[\begin{aligned} \int_3^1 (-2x)\,dx &= -\int_1^3 (-2x)\,dx \\ &= -(-8) = \cb{8} \end{aligned}\]
Practice
Evaluate \(\displaystyle\int_8^2 \dfrac{x}{2}\,dx\) and \(\displaystyle\int_8^8 \dfrac{x}{2}\,dx\).
Drag the orange end from 8 to 2. The region under the line is a trapezoid, and the arrow along the axis shows its sign.
Solution
From 2 to 8, the region under the line is a trapezoid with parallel sides 1 and 4 and width 6.
\[\int_2^8 \tfrac{x}{2}\,dx = \tfrac{1 + 4}{2}\cdot 6 = 15\]The first integral runs from 8 to 2, from right to left, so the red region counts as the negative of that area.
\[\int_8^2 \tfrac{x}{2}\,dx = \cr{-15}\]The second integral has equal bounds, so the region has no width.
\[\int_8^8 \tfrac{x}{2}\,dx = 0\]
Questions
The figure shows the line \(f(x) = \tfrac{x}{2} + \tfrac32\). The shaded region lies between the line and the axis, from \(x = 1\) to \(x = 5\), and its area is 12.
What is \(\displaystyle\int_5^1 f(x)\,dx\)?
- \(-12\)
- 12
- 0
Answer
\(-12\)
What is \(\displaystyle\int_3^3 f(x)\,dx\)?
- 0
- 3
- It does not exist.
Answer
0
What is \(\displaystyle\int_6^2 (10 - x)\,dx\)?
- \(-24\)
- 24
- \(-40\)
Answer
\(-24\)
Summary
If the lower bound is larger than the upper bound, you swap the bounds and change the sign of the integral.
The width of each strip is the end minus the start, so a path from right to left makes every strip count as negative.
If the two bounds are equal, the region has no width, so the integral is 0.
This is one of four skills in topic 6.6 of Which & Why: AP Calculus.