Wemazi

Topic 8.4

Find the horizontal limits of the region

Builds on interpreting an accumulated amount in context. 6 minutes.

Where the curves meet

The line \(y = 2x + 3\) and the parabola \(y = x^2\) enclose the shaded region. No vertical sides are given, so the curves alone show where the region starts and ends.

The orange segment is a vertical strip between the two curves at one x-value. Its length is the height of the line minus the height of the parabola.

Drag the strip across the region. It shrinks to a point at \(x = -1\) and at \(x = 3\), because the curves meet there. At \(x = -1\) both curves are 1 high, and at \(x = 3\) both are 9 high.

Beyond either point, the parabola is above the line, so the strip turns red and its length is negative. The region exists only between the two points, so \(-1\) and \(3\) are its limits, the x-values where it starts and ends.

Any two curves

The curves \(f\) and \(g\) enclose a region. The strip shrinks to a point at the left limit \(a\) and at the right limit \(b\).

At a limit the two curves have the same height, so a limit is an x-value where \(f(x) = g(x)\).

Add the two vertical sides. When a problem gives sides like these, they cut the region, so they are the limits and the meeting points are not.

Notation

  1. The region starts at the left limit \(\cy{a}\) and ends at the right limit \(\cy{b}\), so x runs from \(\cy{a}\) to \(\cy{b}\).

    \[\cy{a} \le x \le \cy{b}\]

  2. If the problem names vertical sides, such as \(x = 1\) and \(x = 4\), those lines are the limits and no equation is needed.

  3. If no sides are given, the limits are where the curves meet. Two curves meet where their heights are equal, so set the two expressions equal.

    \[\cy{f(x)} = \cy{g(x)}\]

  4. Solve the equation, and substitute each solution into both original equations. Squaring can add a false solution, which gives two different heights.

    \[x = \cy{a}, \qquad x = \cy{b}\]

When it fails

Some equations have three solutions. Replace the curves with a line and a cubic. They meet three times, at \(x_1 < x_2 < x_3\), so they enclose two regions.

Each region has its own two limits, \(x_1\) to \(x_2\) and \(x_2\) to \(x_3\). Use neighboring solutions as limits, not only the smallest and the largest.

Example 1: two parabolas

Find the limits of the region enclosed by \(y = 12 - x^2\) and \(y = x^2 - 6\).

  1. The problem names no sides, so the limits are where the curves meet. Set the two heights equal.

    \[12 - x^2 = x^2 - 6\]

  2. Add \(x^2\) and 6 to both sides, which gives \(18 = 2x^2\). Then \(x^2 = 9\), and 9 has two square roots, so \(x = 3\) or \(x = -3\).

    \[18 = 2x^2, \quad x^2 = 9, \quad x = \pm 3\]

  3. Substitute \(x = 3\) into both equations. The first curve is \(12 - 9 = 3\) high, and the second is \(9 - 6 = 3\) high. Both equations use \(x^2\), so \(x = -3\) gives the same heights.

    \[12 - 9 = 3 = 9 - 6\]

  4. The left limit is \(\cy{-3}\), and the right limit is \(\cy{3}\).

    \[\cy{-3 \le x \le 3}\]

Example 2: a square root and a line

Find the limits of the region bounded by \(y = \sqrt{x + 2}\), the line \(y = x\) and the y-axis.

  1. The y-axis is the line \(x = 0\), so the left limit is \(\cy{0}\). The curves give the right limit, so set them equal.

    \[\sqrt{x + 2} = x\]

  2. A tempting wrong move: Square both sides, factor, and keep both solutions as limits.

    \[\begin{aligned} x + 2 &= x^2 \\ (x + 1)(x - 2) &= 0 \\ x = -1 &\text{ or } x = 2 \end{aligned}\]

    At \(x = -1\) the curve is 1 high, but the line is \(-1\) high. The strip there has length 2, so the curves do not meet. Squaring added this false solution, because 1 and \(-1\) have the same square.

  3. Substitute each solution into the original equation, not the squared one. At \(x = 2\) both sides equal 2, but at \(x = -1\) the sides are 1 and \(-1\).

    \[\begin{aligned} \sqrt{2 + 2} &= 2 \\ \sqrt{-1 + 2} &= 1 \ne -1 \end{aligned}\]

  4. The only meeting point is \(x = 2\), so the limits are \(\cy{0}\) and \(\cy{2}\).

    \[\cy{0 \le x \le 2}\]

Practice

Find the limits of the region enclosed by \(y = x^2 - 2x\) and \(y = x + 4\).

Drag the strip across the region. At which two x-values does it shrink to a point?

Solution

  1. Set the two heights equal and move every term to one side.

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

  2. Factor the equation. Each factor gives one solution.

    \[\begin{aligned} (x - 4)(x + 1) &= 0 \\ x = 4 \text{ or } x &= -1 \end{aligned}\]

  3. At \(x = 4\) both curves are 8 high, and at \(x = -1\) both are 3 high, so both solutions are meeting points. The limits are \(\cy{-1}\) and \(\cy{4}\).

    \[\cy{-1 \le x \le 4}\]

Questions

The graph shows the parabola \(y = x^2 + 1\) and the line \(y = 3x - 1\).

  1. Where does the region enclosed by the two curves start and end?

    • \(x = 1\) and \(x = 2\)
    • \(x = 2\) and \(x = 5\)
    • \(x = 0\) and \(x = 3\)
    Answer

    \(x = 1\) and \(x = 2\)

  2. A region is bounded by the two curves and by the line \(x = 3\). What are its limits?

    • \(x = 2\) and \(x = 3\)
    • \(x = 1\) and \(x = 3\)
    • \(x = 1\) and \(x = 2\)
    • \(x = 0\) and \(x = 3\)
    Answer

    \(x = 2\) and \(x = 3\)

  3. Find the limits of the region enclosed by \(y = x^2\) and \(y = x + 6\).

    • \(x = -2\) and \(x = 3\)
    • \(x = -3\) and \(x = 2\)
    • \(x = 4\) and \(x = 9\)
    • \(x = 0\) and \(x = 3\)
    Answer

    \(x = -2\) and \(x = 3\)

Summary

You read a limit from a vertical side that the problem names, or you find it by setting the two heights equal and solving.

This works because the strip between the curves shrinks to a point where they meet, so the region ends there, as at \(x = -1\) and \(x = 3\).

You find the limits before you write an integral for the area between two curves.

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