Wemazi

Topic 6.4

Handle two moving endpoints

Builds on multiplying by the speed of a moving edge. 8 minutes.

One edge adds area and one edge removes it

The line is \(y=f(t)=t+1\). The teal region lies between two moving edges, the lower edge at \(x\) and the upper edge at \(2x\), and its area is \(A(x)\).

At \(x=2\), the edges are at 2 and 4, where the line is 3 high and 5 high.

The upper edge moves 2 units to the right for each unit of \(x\). The line is 5 high there, so the green strip adds area at the rate \(5\cdot2=10\).

The lower edge moves 1 unit to the right for each unit of \(x\). The line is 3 high there, so the red strip removes area at the rate \(3\cdot1=3\).

The net rate, the gain minus the loss, is \(10-3=7\) for each unit of \(x\).

The region is a difference of two areas

The region lies between the lower edge \(l(x)\) and the upper edge \(u(x)\). The faint mark \(c\) is a fixed helper point to the left of both edges.

The area from \(c\) to \(u(x)\) contains the region and the area from \(c\) to \(l(x)\), so the region is their difference.

Each of these areas has one moving edge, so it changes at the rate from the last skill: height times speed at that edge. The second area is subtracted, so the lower edge removes area.

The upper edge adds the green strip, and the lower edge removes the red strip. Narrow the push. The faint line through two nearby points of the lower graph turns onto the orange tangent line. Its slope is the rate of the gain minus the rate of the loss.

Notation

  1. The curve is \(f\). The lower edge is at \(t = \cy{l(x)}\), and the upper edge is at \(t = \cy{u(x)}\). Split the integral at a fixed number \(c\) to the left of both edges.

    \[\begin{aligned} \cb{A(x)} &= \int_{l(x)}^{u(x)} f(t)\,dt \\ &= \int_{c}^{u(x)} f(t)\,dt \\ &\quad - \int_{c}^{l(x)} f(t)\,dt \end{aligned}\]

  2. A push \(\Delta x\) adds a green strip at the upper edge. It is \(f(u(x))\) tall and about \(u'(x)\,\Delta x\) wide, as with one moving edge.

  3. The same push removes a red strip at the lower edge. It is \(f(l(x))\) tall and about \(l'(x)\,\Delta x\) wide.

  4. The derivative is the gain minus the loss, and \(c\) does not appear in it.

    \[\begin{aligned} \frac{dA}{dx} &= \cg{f(u(x))\,u'(x)} \\ &\quad - \cr{f(l(x))\,l'(x)} \end{aligned}\]

    This needs \(f\) continuous on an interval containing both edges, and \(u\) and \(l\) differentiable at \(x\). If the lower edge moves to the left, then \(l'(x)\) is negative, and subtracting its term adds area.

When it fails

The rule needs \(f\) to be continuous at both edges. Put a jump in \(f\) at the faint mark \(p\) and move the upper edge onto it.

Just to the left of \(p\) the strip has one height, and just to the right it has another. The area has no break, but the two red half-lines on its graph have different slopes. \(A\) has a corner there, so \(A'\) does not exist.

Example 1: Edges at x and 4x

Let \(y = \displaystyle\int_x^{4x} e^{-t^2}\,dt\). Find \(\dfrac{dy}{dx}\).

  1. The upper edge \(4x\) moves at speed 4. The lower edge \(x\) moves at speed 1.

  2. The upper edge adds the height at \(4x\) times 4. The lower edge removes the height at \(x\) times 1.

    \[\cg{e^{-(4x)^2}\cdot4}\ \text{ and }\ \cr{e^{-x^2}\cdot1}\]

  3. Subtract the loss from the gain, and write \((4x)^2\) as \(16x^2\).

    \[\dfrac{dy}{dx} = \cg{4e^{-16x^2}} - \cr{e^{-x^2}}\]

Example 2: A lower edge that moves faster than x

Let \(y = \displaystyle\int_{x^2}^{3x} \frac{dt}{1 + t^2}\), for \(0 < x < 3\). Find \(\dfrac{dy}{dx}\) and the value at \(x = 1\).

  1. The upper edge is \(3x\), with speed 3. The lower edge is \(x^2\), with speed \(2x\).

  2. A tempting wrong move: Both edges move to the right, so both terms add area.

    \[\dfrac{dy}{dx} = \dfrac{3}{1+9x^2} + \dfrac{2x}{1+x^4}\]

    The lower edge moves to the right and cuts the red strip off the left end of the region. The area loses that strip, so its term must be subtracted. The red line on the lower graph rises, but the graph of \(y\) falls.

  3. The upper edge adds area and the lower edge removes it, so subtract the lower term.

    \[\dfrac{dy}{dx} = \cg{\dfrac{3}{1+9x^2}} - \cr{\dfrac{2x}{1+x^4}}\]

  4. At \(x = 1\), the edges are at 1 and 3. The gain is \(\frac{3}{10}\), and the loss is \(\frac{2}{2} = 1\), so the area shrinks.

    \[\dfrac{3}{1+9} - \dfrac{2}{1+1} = -\dfrac{7}{10}\]

Practice

Let \(y = \displaystyle\int_{-x}^{2x} e^{t}\,dt\). Find \(\dfrac{dy}{dx}\).

On the figure, the lower edge at \(-x\) moves to the left when \(x\) grows. A region that grows at its left end gains area there.

Solution

  1. The upper edge is \(2x\), with speed 2. The lower edge is \(-x\), with speed \(-1\), because the edge moves to the left.

  2. Subtract the lower term. Its speed \(-1\) is negative, so the subtraction adds area.

    \[\begin{aligned} \dfrac{dy}{dx} &= \cg{e^{2x}\cdot2} - e^{-x}\cdot(-1) \\ &= 2e^{2x} + e^{-x} \end{aligned}\]

  3. As a check, integrate first. The integral is \(e^{2x} - e^{-x}\), and its derivative is the same.

    \[\dfrac{d}{dx}\left(e^{2x} - e^{-x}\right) = 2e^{2x} + e^{-x}\]

Questions

Questions 1 and 2 use the line \(y = f(t) = 10 - t\). The teal region lies between the lower edge at \(x\) and the upper edge at \(x+3\), and its area is \(A(x)\).

  1. At \(x = 2\), how high is the line at the lower edge?

    • 8
    • 5
    • 2
    Answer

    8

  2. At \(x = 2\), what is \(A'(2)\)?

    • \(-3\)
    • 3
    • 13
    Answer

    \(-3\)

  3. Let \(D(x) = \displaystyle\int_x^{x^2} t^2\,dt\). What is \(D'(2)\)?

    • 60
    • 68
    • 64
    Answer

    60

Summary

You differentiate an integral with two moving bounds by adding the upper term, \(f(u)\,u'\), and subtracting the lower term, \(f(l)\,l'\).

This works because the region is the difference of two areas, and each area changes at the rate of height times speed.

You use it when both bounds of an integral depend on \(x\).

This is one of four skills in topic 6.4 of Which & Why: AP Calculus.