Split the area calculation correctly

Where to cut the region

The curve is \(f(x) = x^3 - 5x^2 + 8x\) and the parabola is \(g(x) = x^2\). The region between them runs from \(x = 0\) to \(x = 4\), and the orange line cuts it into two pieces. Drag the cut.

The region is teal where \(f\) is above \(g\) and red where \(f\) is below \(g\).

Move the cut to \(x = 2\). Each piece has one color: the left piece is teal and the right piece is red.

At any other position, one piece has both colors, which means no single curve is on top across that piece. An integral of top minus bottom needs one top curve, so the cut must be at the crossing.

One piece for each stretch

Two curves cross at \(x_1\) and \(x_2\) between the bounds \(a\) and \(b\). The crossings cut the region into three pieces, and each piece has its own top curve and its own limits, such as \(a\) and \(x_1\).

Drag the bar across the region. Its length is top minus bottom, so each piece is labeled \(g - f\) where \(g\) is on top and \(f - g\) where \(f\) is on top. The bar changes color only at a crossing, because inside a piece the top curve stays the same.

Press the button to show the absolute value. Every piece is labeled \(|f - g|\) and the red pieces turn teal, because an absolute value is never negative. One integral of \(|f - g|\) then gives the whole area, but to evaluate it you still need the cuts.

Notation

1. Cut the interval from \(a\) to \(b\) at every crossing inside it. Each piece then has one top curve and one bottom curve.

   \[a < \cy{x_1} < b\]

2. On each piece, integrate top minus bottom. Then add the integrals. Here \(f\) is on top from \(a\) to \(x_1\), and \(g\) is on top from \(x_1\) to \(b\).

   \[\begin{aligned} \cb{A} &= \int_a^{x_1} \left(f - g\right) dx \\ &\quad + \int_{x_1}^{b} \left(g - f\right) dx \end{aligned}\]

3. An equivalent setup is one integral of the absolute value, which is never negative.

   \[\cb{A} = \int_a^b \left|f(x) - g(x)\right| dx\]

4. Where \(f\) is on top, \(f - g\) is positive, so \(|f - g| = f - g\). Where \(g\) is on top, \(|f - g| = g - f\). Both setups give the same area.

   Each setup needs the crossings between \(a\) and \(b\).

When it fails

The integral of \(|f - g|\) looks like one integral, but you cannot evaluate it with one antiderivative, because its formula changes at each crossing.

Press the button to remove the absolute value bars. Use \(f - g\) where it is positive and \(g - f\) where it is negative, the same cuts as the first setup.

Example 1: Two exponentials

Set up the area between \(f(x) = e^x\) and \(g(x) = e^{-x}\) from \(x = -1\) to \(x = 2\). Do not evaluate it.

1. To find the crossing, set the formulas equal. Two powers of \(e\) are equal only when their exponents are equal.

   \[\begin{aligned} e^{x} &= e^{-x} \\ x &= -x \\ x &= 0 \end{aligned}\]

2. The crossing \(x = 0\) lies between the bounds and cuts the region into the pieces from \(-1\) to 0 and from 0 to 2.

3. Test one point in each piece. At \(x = -\tfrac12\), \(e^{-x} = e^{1/2}\) is above \(e^{x} = e^{-1/2}\). At \(x = 1\), \(e^{x} = e\) is above \(e^{-x} = \tfrac1e\).

   \[\begin{aligned} g &\text{ is on top on } [-1, 0] \\ f &\text{ is on top on } [0, 2] \end{aligned}\]

4. Write one integral for each piece, with top minus bottom.

   \[\begin{aligned} \cb{A} &= \int_{-1}^{0} \left(e^{-x} - e^{x}\right) dx \\ &\quad + \int_{0}^{2} \left(e^{x} - e^{-x}\right) dx \end{aligned}\]

Example 2: A cubic and the x-axis

Set up the area between \(f(x) = x^3 - x\) and the x-axis from \(x = -2\) to \(x = 1\). Do not evaluate it.

1. Find the crossings of the curve with the x-axis. Set \(f(x) = 0\) and factor.

   \[\begin{aligned} x^3 - x &= x(x - 1)(x + 1) = 0 \\ x &= -1,\ 0,\ 1 \end{aligned}\]

2. The crossings \(x = -1\) and \(x = 0\) lie inside the bounds, and the crossing \(x = 1\) is the right bound. The pieces run from \(-2\) to \(-1\), from \(-1\) to 0, and from 0 to 1.

3. Test one point in each piece. The curve is below the x-axis at \(x = -1.5\), above it at \(x = -0.5\), and below it at \(x = 0.5\).

   \[\begin{aligned} f(-1.5) &= -1.875 \\ f(-0.5) &= 0.375 \\ f(0.5) &= -0.375 \end{aligned}\]

4. A tempting wrong move: Integrate \(f - g = x^3 - x\) on every piece.

   \[\int_{-2}^{-1} \left(x^3 - x\right) dx = -\tfrac94\]

   On the first piece, the curve is below the x-axis, so \(f - g\) is negative there. Its integral is \(-\tfrac94\), but an area is never negative.

5. Where the curve is below the x-axis, the axis is on top, so top minus bottom is \(0 - (x^3 - x)\).

   \[\begin{aligned} \cb{A} &= \int_{-2}^{-1} \left(0 - (x^3 - x)\right) dx \\ &\quad + \int_{-1}^{0} \left(x^3 - x\right) dx \\ &\quad + \int_{0}^{1} \left(0 - (x^3 - x)\right) dx \end{aligned}\]

Practice

Set up the area between \(f(x) = x^2\) and \(g(x) = 4x - 3\) from \(x = 0\) to \(x = 4\). Do not evaluate it.

Move the cut to where the curves meet. Then test one point in each piece to decide which curve is on top.

Questions

Questions 1 and 2 use \(f(x) = \dfrac{4}{x}\) and \(g(x) = 5 - x\) from \(x = \tfrac12\) to \(x = 5\).

1. Where should the interval from \(\tfrac12\) to 5 be cut?

   • At \(x = 1\) and \(x = 4\)
   • At \(x = 1\) only
   • At \(x = 2.75\), the middle
   • At \(x = \tfrac12\) and \(x = 5\)
   Answer

   At \(x = 1\) and \(x = 4\)

2. Which integral gives the area of the middle piece, from \(x = 1\) to \(x = 4\)?

   • \(\displaystyle\int_1^4 \left[(5 - x) - \tfrac4x\right] dx\)
   • \(\displaystyle\int_1^4 \left[\tfrac4x - (5 - x)\right] dx\)
   • \(\displaystyle\int_{1/2}^{5} \left[(5 - x) - \tfrac4x\right] dx\)
   • \(\displaystyle\int_1^4 \left[(5 - x) + \tfrac4x\right] dx\)
   Answer

   \(\displaystyle\int_1^4 \left[(5 - x) - \tfrac4x\right] dx\)

3. The curves \(f(x) = x\) and \(g(x) = 2x - x^2\) cross at \(x = 0\) and \(x = 1\). Which setup gives the area from \(x = 0\) to \(x = 3\)?

   • \(\displaystyle\int_0^1 (x - x^2)\,dx + \int_1^3 (x^2 - x)\,dx\)
   • \(\displaystyle\int_0^3 (x - x^2)\,dx\)
   • \(\displaystyle\int_0^1 (x - x^2)\,dx + \int_1^3 (x - x^2)\,dx\)
   • \(\displaystyle\int_0^{3/2} (x - x^2)\,dx + \int_{3/2}^3 (x^2 - x)\,dx\)
   Answer

   \(\displaystyle\int_0^1 (x - x^2)\,dx + \int_1^3 (x^2 - x)\,dx\)

Summary

You cut the region at every crossing and integrate top minus bottom on each piece.

This works because each piece has one top curve, so top minus bottom is positive on the whole piece.

Use it when the curves cross between the bounds.

This is one of four skills in topic 8.6 of Which & Why: AP Calculus.