Wemazi

Topic 6.4

Differentiate an accumulation with a variable endpoint

Builds on finding where the accumulation increases or decreases. 6 minutes.

The edge moves and the area grows

The upper panel shows the curve \(y=f(t)=\frac{t^2}{2}\). The teal region under it, from 0 to the orange edge at \(x\), has area \(F(x)\), and the lower panel plots \(F(x)\) against \(x\).

If the edge at \(x=3\) moves to the right by \(h=0.1\), the region gains a thin green strip. The strip is almost a rectangle 0.1 wide and \(f(3)=4.5\) tall, so its area is about 0.45.

The exact area of the strip is 0.465. Divide it by \(h\) to find how fast the area grows for each unit of \(x\), which gives 4.652.

Make \(h\) smaller. For \(h=0.01\) the result is 4.515, and the results approach 4.5, which is \(f(3)\).

The area grows at the rate of the height of the curve at the moving edge.

The rate of growth is the height at the edge

The fixed left edge is \(a\), and the moving right edge is \(x\). The area between them is \(F(x)\).

Move the edge from \(x\) to \(x+h\). The area changes by the strip's area, \(F(x+h)-F(x)\). Over the strip the curve has a lowest and a highest height, so the strip's area lies between those heights times \(h\).

If \(f\) is continuous, both heights approach \(f(x)\) as \(h\) shrinks, so the rate \(\frac{F(x+h)-F(x)}{h}\) approaches \(f(x)\). Narrow the strip. The faint line through two nearby points of the lower graph turns onto the orange tangent line.

Move the slider for \(a\). The lower graph moves up or down, but the slope of the tangent does not change. The start changes the amount, not the rate.

Notation

  1. The curve is \(f\), and its input is \(t\), because \(x\) names the edge. The region runs from the fixed start \(t = a\) to the edge at \(t = \cy{x}\).

  2. The area depends on the edge, so it is a function of \(x\). Call it \(\cb{F}\).

    \[\cb{F(x) = \int_a^x f(t)\,dt}\]

  3. The height of the curve at the edge, \(\cy{f(x)}\), is the slope of the graph of \(\cb{F}\).

  4. The derivative of the integral from \(a\) to \(x\) of \(f(t)\,dt\) is \(f(x)\). The answer is a height, not an area, and it does not contain \(a\).

    \[\frac{d}{dx}\,\cb{\int_a^x f(t)\,dt} = \cy{f(x)}\]

    This needs \(f\) continuous on an interval containing \(a\) and \(x\). You do not need an antiderivative of \(f\).

When it fails

The rule needs \(f\) to be continuous at \(x\). Put a jump in \(f\) and move the edge onto it.

Left of the jump at \(c\) the curve has one height, and right of it another. The area has no break, but its slope is different on the two sides, so \(F\) has a corner and \(F'(c)\) does not exist.

Example 1: A curve with no simple antiderivative

Let \(G(x) = \displaystyle\int_1^x \sqrt{t^3 + 7}\,dt\). Find \(G'(x)\) and \(G'(2)\).

  1. The lower bound 1 is fixed, and the upper bound \(\cy{x}\) moves. The integrand has no simple antiderivative, so use the rule: each new strip is as tall as the curve at the edge.

  2. Replace \(t\) by \(x\) in the integrand.

    \[G'(x) = \cy{\sqrt{x^3 + 7}}\]

  3. At \(x = 2\), the derivative is the height of the curve at the edge, and the tangent in the lower graph has this slope.

    \[G'(2) = \cy{\sqrt{2^3 + 7}} = \sqrt{15}\]

    As a check, \(\sqrt{15} \approx 3.87\).

Example 2: The moving edge is the lower bound

Let \(H(x) = \displaystyle\int_x^4 \sqrt{t^2 + 9}\,dt\). Find \(H'(x)\) and \(H'(0)\).

  1. The upper bound 4 is fixed, and the lower bound \(\cy{x}\) moves. The region lies between them.

  2. A tempting wrong move: Replace \(t\) by \(x\) in the integrand, as before.

    \[H'(x) = \sqrt{x^2 + 9}\]

    When \(x\) moves to the right, the region loses the red strip at its left edge. The area shrinks, so \(H\) falls, but this positive answer is the slope of the red line, which rises.

  3. The rule needs the moving bound on top. Swapping the bounds changes the sign of the integral, so the lost strip counts as negative.

    \[\begin{aligned} H(x) &= -\int_4^x \sqrt{t^2 + 9}\,dt \\ H'(x) &= \cy{-\sqrt{x^2 + 9}} \end{aligned}\]

  4. At \(x = 0\), the region loses a strip of height 3 for each unit that the edge moves.

    \[H'(0) = -\sqrt{0^2 + 9} = \cy{-3}\]

Practice

Let \(A(x) = \displaystyle\int_{-2}^{x} \frac{1}{1 + t^4}\,dt\). Find \(A'(1)\).

Drag the edge to the right. The area grows at the rate of the height of the curve at the edge, so read that height at x = 1.

Solution

  1. The moving edge is the upper bound, so replace \(t\) by \(x\) in the integrand.

    \[A'(x) = \dfrac{1}{1 + x^4}\]

  2. At \(x = 1\), the derivative is the height of the curve at the edge.

    \[A'(1) = \dfrac{1}{1 + 1^4} = \cy{\dfrac12}\]

Questions

Questions 1 and 2 use the line \(y = f(t) = 6 - 2t\) and the amount \(F(x) = \int_0^x f(t)\,dt\). The region from 0 to \(x\) is teal above the axis and red below it, where it counts as negative.

  1. At \(x = 1\), how fast does \(F\) change with respect to \(x\)?

    • 4
    • 5
    • 6
    Answer

    4

  2. At \(x = 4\), what is \(F'(4)\)?

    • \(-2\)
    • 8
    • 2
    Answer

    \(-2\)

  3. Let \(R(x) = \displaystyle\int_3^x \frac{t^2}{t + 1}\,dt\). What is \(R'(5)\)?

    • \(\dfrac{25}{6}\)
    • \(\dfrac{9}{4}\)
    • \(-\dfrac{25}{6}\)
    Answer

    \(\dfrac{25}{6}\)

Summary

You differentiate an integral with a moving upper bound by writing the integrand at that bound.

This works because a small move of the edge adds a thin strip, and its area is the height at the edge times its width.

You use it to find the rate of an accumulated amount, even when the integrand has no simple antiderivative.

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