Set up an integral with respect to x

Thin strips between two curves

The parabola \(y = 3x - x^2\) and the line \(y = x\) enclose a region from \(x = 0\) to \(x = 2\), and the parabola is the upper curve.

Move the slider to cut the region into thin vertical strips. Each strip is almost a rectangle, and its width \(\Delta x\) is 2 divided by the number of strips.

A strip's height is the upper height minus the lower height, \((3x - x^2) - x\), at its middle. Its area is this height times its width. The orange strip shows its height beside it and its green width \(\Delta x\) under it.

With 4 strips the areas add to 1.375, and with 8 strips they add to about 1.344. As the strips become thinner, the sum approaches the exact area, \(\tfrac{4}{3}\).

Any two curves

The upper curve is \(f\), the lower curve is \(g\), and the region lies between \(x = a\) and \(x = b\).

Move the strip. At every \(x\), its height is \(f(x) - g(x)\), the upper curve minus the lower curve. Its area is this height times its width \(dx\).

Make the strips thinner. Their total changes less and less, and the integral from \(a\) to \(b\) is the value it approaches, the area of the region.

Notation

1. A thin strip at \(x\) is \(\cy{f(x) - g(x)}\) tall and \(\cg{dx}\) wide, so its area is its height times its width.

   \[\text{strip area} = \cy{\big(f(x) - g(x)\big)}\,\cg{dx}\]

2. Add the areas of the strips from the left limit \(\cy{a}\) to the right limit \(\cy{b}\). As the strips become thinner, their total approaches the area \(\cb{A}\), and the integral is that value.

   \[\cb{A} = \cb{\int_{\cy{a}}^{\cy{b}}} \cy{\big(f(x) - g(x)\big)}\,\cg{dx}\]

3. The integrand, the expression that you integrate, is the upper curve minus the lower curve. The limits are x-values, not heights.

   This needs \(f\) and \(g\) continuous on \([a, b]\), with \(f(x) \ge g(x)\) there.

4. If the upper curve or the lower curve changes formula, split the region. Write one integral for each piece, with its own limits.

   \[\begin{aligned} &\int_a^c \big(f_1 - g_1\big)\,dx \\ &\quad + \int_c^b \big(f_2 - g_2\big)\,dx \end{aligned}\]

When it fails

The integral needs \(f\) above \(g\) from \(a\) to \(b\). Push the right limit \(b\) past the point where the curves meet.

Past that point, \(g\) is on top, so \(f(x) - g(x)\) is negative and the piece turns red. The integral subtracts this piece instead of adding it, so split the region where the curves meet.

Example 1: an exponential curve and a line

Set up the integral for the area of the region between \(y = e^x\) and \(y = 1\), from \(x = 0\) to \(x = \ln 3\).

1. The statement gives the sides, \(x = 0\) and \(x = \ln 3\), so these are the limits.

   \[0 \le x \le \ln 3\]

2. Test \(x = 0.5\), which lies between the limits. The curve \(e^x\) is about 1.65 high and the line is 1 high, so \(e^x\) is the upper curve.

3. Subtract the lower height from the upper height, and integrate between the limits.

   \[\cb{\int_0^{\ln 3}} \big(e^x - 1\big)\,dx\]

   As a check, this integral equals \(2 - \ln 3\), about 0.90, which is positive as an area must be.

Example 2: a parabola and a line

Set up the integral for the area of the region enclosed by \(y = 10 - x^2\) and \(y = x + 4\).

1. Set the heights equal to find the limits. The solutions are \(x = -3\) and \(x = 2\).

   \[\begin{aligned} 10 - x^2 &= x + 4 \\ x^2 + x - 6 &= (x + 3)(x - 2) = 0 \end{aligned}\]

2. At \(x = 0\) the parabola is 10 high and the line is 4 high, so the parabola is the upper curve.

3. A tempting wrong move: Integrate the upper curve alone.

   \[\int_{-3}^{2} (10 - x^2)\,dx\]

   This is the area between the parabola and the x-axis, shown in red. It includes the part below the line, but the strips of the region stop at the line.

4. Subtract the lower height from the upper height.

   \[\begin{aligned} &\cb{\int_{-3}^{2}} \big[(10 - x^2) - (x + 4)\big]\,dx \\ &\quad = \int_{-3}^{2} (6 - x - x^2)\,dx \end{aligned}\]

Practice

Set up the integral for the area of the region enclosed by \(y = 4 - x^2\) and \(y = 3x\).

Drag the strip to x = 0 and read the two heights. Which curve is the upper curve?

Questions

The graph shows the parabola \(y = 6x - x^2\) and the line \(y = 2x\).

1. Which integral gives the area of the region between the curves?

   • \(\displaystyle\int_0^4 (4x - x^2)\,dx\)
   • \(\displaystyle\int_0^4 (x^2 - 4x)\,dx\)
   • \(\displaystyle\int_0^4 (6x - x^2)\,dx\)
   • \(\displaystyle\int_0^8 (4x - x^2)\,dx\)
   Answer

   \(\displaystyle\int_0^4 (4x - x^2)\,dx\)

2. The region is cut into four strips of width 1. Each strip is as tall as the gap between the curves at its middle. What is the area of the second strip, from \(x = 1\) to \(x = 2\)?

   • 3.75
   • 3
   • 6.75
   • 1
   Answer

   3.75

3. Which integral gives the area of the region enclosed by \(y = 4\) and \(y = x^2\)?

   • \(\displaystyle\int_{-2}^{2} (4 - x^2)\,dx\)
   • \(\displaystyle\int_{-2}^{2} (x^2 - 4)\,dx\)
   • \(\displaystyle\int_0^4 (4 - x^2)\,dx\)
   • \(\displaystyle\int_{-2}^{2} (4 + x^2)\,dx\)
   Answer

   \(\displaystyle\int_{-2}^{2} (4 - x^2)\,dx\)

Summary

You write the integral of the upper curve minus the lower curve, with the x-values of the left and right sides as the limits.

This works because each thin strip has an area of height times width, and the integral adds the areas of all the strips.

You use it when a region lies between two curves and the strips stand vertically.

This is one of four skills in topic 8.4 of Which & Why: AP Calculus.