Calculus 1

Graphing an accumulation function

You know how fast an amount is changing at every moment. When is the amount largest, how large is it, and what does its whole graph look like?

You need

The first and second derivative testsCalculus 1 Signed area and integrals from geometryCalculus 1

Where you are 3 in Accumulation and net change

Calculus8

Applications of integrals10

Accumulation and net change3

Graphing an accumulation function

Position, displacement and distance from velocity The net change theorem in context

When you reach for it

Rauf, Togrul and Lale work the doors of an indoor arena. Turnstiles count people coming in, other counters count people going out, and the difference is a net rate: people per hour entering minus people per hour leaving. The rate is what they can measure; the number of people inside is what they care about. If r(t) is the net rate and P_0 people were inside at time 0, the number inside at time x is

P(x) = P_0 + \int_0^x r(t)\,dt.

You reach for this move whenever you hold the graph or formula of a rate and want the shape of the amount it builds: where it rises, falls and bends, where it peaks and how high. The graph you hold is the derivative of the function you want, so the reading is the one from Curve sketching and connecting f, f' and f'', where f was read from a graph of f'; what is new is that heights now come exactly, from area. The crowd is largest when the net rate crosses from positive to negative.

The procedure

Day 1: reading g from the graph of f

Throughout, f is continuous on [a, b] and

g(x) = \int_a^x f(t)\,dt, \qquad a \le x \le b.

Heights come from area. First, g(a) = 0. For any two inputs, Properties of definite integrals gives

g(x_2) - g(x_1) = \int_{x_1}^{x_2} f(t)\,dt,

the signed area between the graph of f and the t-axis from x_1 to x_2 (Signed area and integrals from geometry). Area above the axis raises g; area below lowers it. When f is built from segments and circular arcs, every piece is a triangle, a trapezoid, a rectangle or part of a disk, so every value of g is exact.

Slopes come from heights of f. By the first fundamental theorem of calculus, g'(x) = f(x) for every x in (a, b): the height of the graph you hold is the slope of the graph you want. Where f > 0, g is increasing, and where f < 0, g is decreasing (Increasing, decreasing, constant, positive and negative intervals; this is the mean value theorem, The mean value theorem). Because g' exists everywhere in (a, b), the critical points of g are exactly the zeros of f there. By the The first and second derivative tests, g has a local maximum at a zero where f changes from positive to negative and a local minimum at a zero where f changes from negative to positive. A zero where f touches the axis without changing sign gives a horizontal tangent and no extremum. As in Critical points and local versus absolute extrema, a local extremum sits at an interior point, so a and b are never local extrema, although either may be an absolute one.

Bending comes from slopes of f. The graph of g is concave up where g' is increasing (Concavity and inflection points), and g' = f, so g is concave up where f is increasing and concave down where f is decreasing. Where f is differentiable this says g'' = f', but the reading needs only whether f rises or falls, so it works across a corner of f too. An inflection point of g sits wherever f switches between rising and falling: at a peak or a valley of f.

On the graph of f	On the graph of g
above the axis	rising
below the axis	falling
crosses the axis from + to -	local maximum
crosses the axis from - to +	local minimum
rising	concave up
falling	concave down
a peak or a valley	inflection point
height f(x)	slope g'(x)
area from x_1 to x_2	change g(x_2) - g(x_1)

Day 2: absolute extrema, context and other lower limits

Absolute extrema. The function g is continuous on [a, b] and differentiable inside, so the Extrema on a closed interval applies: its absolute maximum and minimum are among its values at a, at b, and at the zeros of f in between. Values are compared by area: g(c_2) > g(c_1) exactly when \int_{c_1}^{c_2} f(t)\,dt > 0, and often that sign can be settled by comparing regions without finding either value. An endpoint can win even when f is not zero there.

The amount in context. If r is the rate of change of an amount Q, then Q(x) = Q(a) + \int_a^x r(t)\,dt, which is g plus a constant. It has the same rises, falls, bends and turning points as g; the constant moves only the values. The units of Q are the units of r times the units of t (The net change theorem in context).

A lower limit other than the left end. For c in [a, b] let h(x) = \int_c^x f(t)\,dt. Additivity gives

\begin{aligned} h(x) &= \int_a^x f(t)\,dt - \int_a^c f(t)\,dt \\ &= g(x) - g(c), \end{aligned}

so the graph of h is the graph of g moved up or down until it passes through (c, 0). Every critical point, extreme point and inflection point stays at the same x. For x < c the integral runs backward, h(x) = -\int_x^c f(t)\,dt, so where f is positive to the left of c, h is negative: it climbs to 0 at c.

Sketching. A complete sketch of g plots exact values at the zeros, corners, peaks and valleys of f and at the ends, with a horizontal tangent at each zero of f and the steepest rise or fall at each peak or valley of f. Where f is constant, g is a straight segment with that slope.

Worked examples

Example 1

Day 1: Rauf finds values by area

The graph of f is shown below: a segment from (0, -2) to (2, 2), a segment from (2, 2) to (4, 0), and on [4, 8] the lower half circle \displaystyle f(t) = -\sqrt{4 - (t - 6)^2}. Rauf sets g(x) = \int_0^x f(t)\,dt. Find g(1), g(2), g(4), g(6) and g(8).

Figure 1The graph of f for Examples 1, 2, 3, 5 and 7. On [4, 8] it is the lower half circle \displaystyle f(t) = -\sqrt{4 - (t - 6)^2}.

Solution

The first segment has slope 2 and passes through (0, -2), so it meets the axis at t = 1. The other zeros are t = 4 and t = 8. Add the signed area one piece at a time, starting from g(0) = 0.

On [0, 1] the region is a triangle below the axis with base 1 and height 2: area 1, counted negative, so g(1) = -1. On [1, 2] a triangle above the axis, base 1, height 2, adds 1: g(2) = 0. On [2, 4] a triangle above, base 2, height 2, adds 2: g(4) = 2. On [4, 6] the region is a quarter of a disk of radius 2, below the axis, with area \tfrac{1}{4}\pi(2)^2 = \pi, so g(6) = 2 - \pi. The second quarter disk on [6, 8] gives g(8) = 2 - 2\pi \approx -4.28. ### Day 1: where Rauf’s g rises, falls and turns

For the same g, find where g is increasing and decreasing, and find its local extrema.

Solution

The pieces of f meet at (2, 2) and (4, 0), so f is continuous on [0, 8] and g'(x) = f(x) by the first fundamental theorem. Read the sign of f off the graph: f < 0 on (0, 1), f > 0 on (1, 4), f < 0 on (4, 8). Hence g is decreasing on [0, 1], increasing on [1, 4] and decreasing on [4, 8].

The critical points in (0, 8) are the zeros 1 and 4. At x = 1, f changes from negative to positive, so g has a local minimum g(1) = -1. At x = 4, f changes from positive to negative, so g has a local maximum g(4) = 2. Although f(8) = 0, the point x = 8 is an endpoint and not a local extremum.

Example 2

Day 1: how Rauf’s g bends, and the sketch

For the same g, find where the graph is concave up and concave down, find the inflection points, and sketch g.

Solution

The graph of g is concave up where g' = f is increasing. On (0, 2), f rises with slope 2; on (2, 4) it falls with slope -1; on (4, 6) the lower half circle falls from 0 to -2; on (6, 8) it rises back to 0. So g is concave up on (0, 2), concave down on (2, 6) and concave up on (6, 8).

The concavity changes at the peak and the valley of f, so the inflection points are (2, 0) and (6, 2 - \pi). At x = 2 the graph of f has a corner (slope 2 on the left, -1 on the right), so g''(2) does not exist; the point is still an inflection point, because g' changes from increasing to decreasing there.

For the sketch, the slopes at the marked points are heights of f: g'(0) = -2, horizontal tangents at x = 1 and x = 4, g'(2) = 2 (the steepest rise on the interval, since 2 is the largest value of f), g'(6) = -2 (the steepest fall), and g'(8) = 0, so the graph levels off at the right end.

Figure 2The graph of g(x) = \int_0^x f(t)\,dt. Lows and highs sit at the zeros of f, bends change at the peak and valley of f.

Example 3

Day 1: an integrand with no area formula

Rauf meets

g(x) = \int_0^x \frac{(t - 2)^2(4 - t)}{t^2 + 1}\,dt, \qquad 0 \le x \le 6.

No region here has a geometric area formula. Find where g is increasing and decreasing, its local extrema, and what the graph of g does at x = 2.

Solution

By the first fundamental theorem,

g'(x) = f(x) = \frac{(x - 2)^2(4 - x)}{x^2 + 1}.

The denominator is positive. The factor (x - 2)^2 is positive except at x = 2, where it is zero, and 4 - x is positive for x < 4 and negative for x > 4. So f > 0 on (0, 2) and on (2, 4), and f < 0 on (4, 6). Then g is increasing on [0, 2] and on [2, 4], hence on all of [0, 4], and decreasing on [4, 6]. The only local extremum is a local maximum at x = 4, where f changes from positive to negative. At x = 2 the tangent is horizontal, but g rises on both sides, so there is no extremum.

For the bending at x = 2, write f(x) = (x - 2)^2 q(x) with q(x) = \dfrac{4 - x}{x^2 + 1}. The product rule gives

f'(x) = (x - 2)\big[\,2q(x) + (x - 2)q'(x)\,\big].

The bracket tends to 2q(2) = \tfrac{4}{5} > 0 as x \to 2, so near x = 2 the sign of f'(x) is the sign of x - 2: f falls just left of 2 and rises just right of it. So g is concave down just left of 2 and concave up just right: (2, g(2)) is an inflection point with a horizontal tangent. The second derivative test could not have told us this, since g''(2) = f'(2) = 0.

Example 4

Day 2: Togrul’s absolute extrema

Togrul takes Rauf’s g(x) = \int_0^x f(t)\,dt from Example 1. Find the absolute maximum and minimum of g (a) on [0, 8] and (b) on [0, 5].

Solution

(a) The candidates are the endpoints 0 and 8 and the zeros of f inside, 1 and 4. From Example 1, g(0) = 0, g(1) = -1, g(4) = 2 and g(8) = 2 - 2\pi \approx -4.28. The absolute maximum is 2, at x = 4. The absolute minimum is 2 - 2\pi, at the endpoint x = 8, which is not a local minimum.

(b) The candidates are 0, 1, 4 and 5. The exact value of g(5) is not needed. Since f < 0 on (4, 5), g(5) < g(4) = 2. And

\begin{aligned} g(5) - g(1) &= \int_1^5 f(t)\,dt \\ &= 1 + 2 - (\text{area of the strip}). \end{aligned}

The strip, the part of the half disk over [4, 5], lies inside the rectangle 4 \le t \le 5, -2 \le y \le 0, of area 2, so its area is less than 2 and g(5) - g(1) > 1 > 0. Every candidate other than x = 1 gives more than -1, so on [0, 5] the absolute maximum is 2 at x = 4 and the absolute minimum is -1 at x = 1, an interior point this time.

Example 5

Day 2: Lale’s crowd

Lale’s net rate r(t) for the arena, in hundreds of people per hour, t hours after 3 pm, is graphed below: a segment from (0, 4) to (1, 8), a segment from (1, 8) to (3, 0), on [3, 7] the dip \displaystyle r(t) = -\sqrt{4 - (t - 5)^2}, and a segment from (7, 0) to (8, 2), where a late show starts. At 3 pm there were 300 people inside. When during the eight hours were the most people inside, and how many? When were the fewest? When was the crowd growing fastest?

Figure 3Lale’s net rate r(t), in hundreds of people per hour, t hours after 3 pm.

Solution

In hundreds of people, P(x) = 3 + \int_0^x r(t)\,dt; hundreds of people per hour times hours gives hundreds of people. Since P' = r, the interior candidates are the zeros of r: t = 3 and t = 7. With the endpoints:

P(0) = 3.
On [0, 1] a trapezoid with parallel sides 4 and 8 and width 1 adds 6; on [1, 3] a triangle with base 2 and height 8 adds 8. So P(3) = 17.
On [3, 7] the region below the axis is half of a disk of radius 2, area 2\pi, so P(7) = 17 - 2\pi \approx 10.72.
On [7, 8] a triangle with base 1 and height 2 adds 1, so P(8) = 18 - 2\pi \approx 11.72.

The most people inside was 1{,}700, at t = 3, that is 6 pm, the moment r crosses from positive to negative. The fewest was 300, at 3 pm. At 10 pm there is a local minimum of about 1{,}072 people, but it is not the fewest, and the late show brings the count back up to about 1{,}172 by 11 pm.

The crowd grows fastest where P' = r is largest: at t = 1, 4 pm, at 800 people per hour. That is an inflection point of P, since r has its peak there.

Example 6

Day 2: Togrul moves the lower limit

With f from Example 1, let h(x) = \int_4^x f(t)\,dt on [0, 8]. Find h(0), h(1), h(2), h(4) and h(8), the absolute extrema of h, and describe its graph.

Solution

By additivity, h(x) = g(x) - g(4) = g(x) - 2, with g from Example 1. So h(0) = -2, h(1) = -3, h(2) = -2, h(4) = 0 and h(8) = -2\pi.

Reading h(2) straight from the graph shows why values to the left of 4 come out negative: h(2) = \int_4^2 f(t)\,dt = -\int_2^4 f(t)\,dt = -2. The triangle between 2 and 4 lies above the axis, but the integral runs from right to left, so it counts against h.

Since h' = f as well, every feature from Examples 2 and 3 sits at the same x, and the graph of h is the graph of g lowered by 2. The absolute maximum is 0 at x = 4 and the absolute minimum is -2\pi at x = 8, so h(x) \le 0 on all of [0, 8].

Boundaries

Continuity of f. The equation g' = f holds at every point only when f is continuous. At a jump of f, g is continuous but has a corner and no derivative (The first fundamental theorem of calculus); the jump joins the list of critical points, and the first derivative test still applies there.
A corner of f leaves g'' undefined, but concavity is still read from whether f rises or falls (Example 3).
A zero without a sign change gives a horizontal tangent, not an extremum (Example 4).
A closed interval is needed for absolute extrema to exist: on all real numbers, \int_0^x 1\,dt = x (a rectangle’s signed area) has no maximum and no minimum.
Exact values need exact areas. Without them the signs of f and f' still give the shape (Example 4), but heights wait for The second fundamental theorem of calculus or are bounded as in Example 5.

Where this goes next

The same reading turns velocity into position (Position, displacement and distance from velocity) and any rate into its total (The net change theorem in context). The second fundamental theorem (The second fundamental theorem of calculus) supplies exact values of g whenever an antiderivative of f is known, and together the two halves make up the fundamental theorem of calculus.

Exercises

Exercises 1 to 5, 11, 12 and 16 to 19 use the graph of k below and A(x) = \int_0^x k(t)\,dt on [0, 9]. The graph is a quarter circle \displaystyle k(t) = \sqrt{4 - t^2} on [0, 2], a segment from (2, 0) to (4, -2), a segment from (4, -2) to (7, 1), and k(t) = 1 on [7, 9].

Figure 4The graph of k for the exercises. On [0, 2] it is the quarter circle \displaystyle k(t) = \sqrt{4 - t^2}.

Day 1: values and features from a graph

Find A(2), A(4), A(6) and A(9).
Find A'(1), A''(1), A'(5) and A''(5).
On what intervals is A increasing, and on what intervals decreasing? Find each local extremum of A and its value.
On what intervals is the graph of A concave up, concave down, or straight? Find every inflection point.
Sketch the graph of A on [0, 9], marking exact values at x = 0, 2, 4, 6, 7, 9 and the slopes there.

Day 1: integrands given by formulas

Let g(x) = \int_0^x (t - 1)(t - 4)e^{-t}\,dt for all real x. Find the local extrema of g and the intervals where its graph is concave up.
Let \displaystyle F(x) = \int_1^x \frac{\cos t}{t}\,dt for 0 < x < 2\pi. Find where F has local maxima and local minima.
Let \displaystyle G(x) = \int_0^x \frac{t}{1 + t^4}\,dt for all real x. Find where G increases and decreases, its extrema, and its inflection points.

Day 1: reasoning

Explain why the graph of g(x) = \int_a^x f(t)\,dt has a horizontal tangent at every zero of a continuous f, and why that alone does not make an extremum. Give an integrand showing it.
Suppose f is continuous, rises on (c - \delta, c] and falls on [c, c + \delta) for some \delta > 0, possibly with a corner at c. Explain why g(x) = \int_a^x f(t)\,dt has an inflection point at c.

Day 2: absolute extrema

Find the absolute maximum and minimum of A on [0, 9].
Find the absolute maximum and minimum of A on [3, 7].
Let F(x) = \int_0^x \big(|t - 3| - 1\big)\,dt on [0, 6]. Find the local and absolute extrema of F. Is either local extremum an absolute one?

Day 2: the amount in context

On a Saturday, the net rate of people entering Rauf’s arena is r(t) = 6 - 2t hundred people per hour, t hours after 1 pm, for 0 \le t \le 6. At 1 pm 400 people were inside. When were the most people inside, and how many? When were the fewest?
Use Lale’s graph from Example 6. How many people were inside at 4 pm and at 8 pm (to the nearest person)? When was the crowd shrinking fastest, and at what rate?

Day 2: a different lower limit

Let B(x) = \int_6^x k(t)\,dt. Find B(0), B(2), B(4) and B(9).
Find the absolute maximum and minimum of B on [0, 9].
Describe how the graph of B is related to the graph of A, and sketch it.

Day 2: reasoning

Without using a formula for B, explain from the graph of k alone why B(x) \ge 0 for every x in [0, 9].
Lale says the arena is most crowded at the moment people are pouring in fastest. Explain why that is wrong, using the numbers from Example 6.
Show that for any a and c, \int_c^x f(t)\,dt - \int_a^x f(t)\,dt does not depend on x, and explain why both functions have their absolute maximum on an interval at the same x.
Suppose f is continuous on [a, b] and f(t) > 0 except at finitely many points, where it is 0. Explain why g(x) = \int_a^x f(t)\,dt has its absolute minimum at a and its absolute maximum at b.

Answers

A(2) = \pi, A(4) = \pi - 2, A(6) = \pi - 4, A(9) = \pi - \tfrac{3}{2}. A quarter disk of radius 2 has area \pi; each triangle on [2, 4] and [4, 6] has area 2 below the axis; then the triangle on [6, 7] adds \tfrac{1}{2} and the rectangle on [7, 9] adds 2.
\displaystyle A'(1) = k(1) = \sqrt{3}; \displaystyle A''(1) = k'(1) = -\dfrac{1}{\sqrt{3}} = -\dfrac{\sqrt{3}}{3}, since \displaystyle k'(t) = -\dfrac{t}{\sqrt{4 - t^2}}; A'(5) = k(5) = -1; A''(5) = 1, the slope of the segment through (4, -2) and (7, 1).
Increasing on [0, 2] and [6, 9], decreasing on [2, 6]. Local maximum A(2) = \pi (k changes from + to -); local minimum A(6) = \pi - 4 (k changes from - to +).
Concave down on (0, 4), where k falls; concave up on (4, 7), where k rises; straight on (7, 9), where k = 1 and A is a segment of slope 1. The only inflection point is (4, \pi - 2), at the valley of k. At x = 7 the graph goes from concave up to straight, which is not a change between up and down.
The points are (0, 0), (2, \pi), (4, \pi - 2), (6, \pi - 4), (7, \pi - \tfrac{7}{2}), (9, \pi - \tfrac{3}{2}), with slopes 2, 0, -2, 0, 1, 1: a peak at x = 2, the steepest fall at x = 4, a low point at x = 6, and a straight segment of slope 1 from x = 7.
g'(x) = (x - 1)(x - 4)e^{-x}, which changes from + to - at 1 and from - to + at 4: a local maximum at x = 1 and a local minimum at x = 4. Then g''(x) = (-x^2 + 7x - 9)e^{-x}, positive between its zeros, so the graph is concave up on \displaystyle \left(\frac{7 - \sqrt{13}}{2}, \frac{7 + \sqrt{13}}{2}\right), with inflection points at both ends of that interval.
F'(x) = \dfrac{\cos x}{x}, with the sign of \cos x since x > 0. It changes from + to - at \displaystyle x = \frac{\pi}{2} (local maximum) and from - to + at \displaystyle x = \frac{3\pi}{2} (local minimum).
G'(x) = \dfrac{x}{1 + x^4}, negative for x < 0 and positive for x > 0: G decreases on (-\infty, 0], increases on [0, \infty), and has its absolute minimum G(0) = 0. G''(x) = \dfrac{1 - 3x^4}{(1 + x^4)^2} changes sign at x = \pm 3^{-1/4}, the inflection points; the graph is concave up between them and concave down outside.
By the first fundamental theorem g'(c) = f(c) = 0, so the tangent is horizontal. An extremum needs g' to change sign, which f need not do: f(t) = t^2 gives g(x) = \int_0^x t^2\,dt, with g'(0) = 0, but g' = x^2 \ge 0 so g is increasing through 0.
Since g' = f, g' is increasing on (c - \delta, c] and decreasing on [c, c + \delta), so g is concave up left of c and concave down right of it: an inflection point. The argument never uses g''(c), so a corner of f at c does not matter.
Candidates 0, 2, 6, 9 give 0, \pi, \pi - 4, \pi - \tfrac{3}{2}. Absolute maximum \pi at x = 2; absolute minimum \pi - 4 at x = 6.
Candidates 3, 6, 7. A(3) = \pi - \tfrac{1}{2} (triangle of base 1 and height 1 below the axis on [2, 3]), A(6) = \pi - 4, A(7) = \pi - \tfrac{7}{2}. Absolute maximum \pi - \tfrac{1}{2} at the endpoint x = 3; absolute minimum \pi - 4 at x = 6.
The integrand is 2 at t = 0, 0 at t = 2, -1 at t = 3, 0 at t = 4, 2 at t = 6, so F has a local maximum F(2) = 2 and a local minimum F(4) = 2 - 1 = 1, and F(6) = 1 + 2 = 3, F(0) = 0. Absolute maximum 3 at x = 6, absolute minimum 0 at x = 0; neither local extremum is absolute.
r changes from + to - at t = 3. In hundreds, P(3) = 4 + 9 = 13 (a triangle with base 3, height 6) and P(6) = 13 - 9 = 4. The most was 1{,}300 people, at 4 pm; the fewest was 400, at both 1 pm and 7 pm.
At 4 pm: 300 + 600 = 900 people. At 8 pm (t = 5): half of the dip, a quarter disk of area \pi, has been subtracted, so P(5) = 17 - \pi \approx 13.858 hundred, about 1{,}386 people. The crowd shrinks fastest where r is lowest: at 8 pm, 200 people per hour.
B(x) = A(x) - A(6) = A(x) - (\pi - 4): B(0) = 4 - \pi, B(2) = 4, B(4) = 2, B(9) = \tfrac{5}{2}.
Absolute maximum 4 at x = 2; absolute minimum 0 at x = 6, the same x values as for A in Exercise 11.
B = A - (\pi - 4), so the graph of B is the graph of A raised by 4 - \pi \approx 0.86; it touches the axis at its low point (6, 0) and has the same shape as the sketch in Exercise 5.
For x > 6, k > 0 on (6, x), so B(x) > 0. For 2 \le x < 6, B(x) = -\int_x^6 k(t)\,dt with k \le 0 on [x, 6], so B(x) \ge 0. For 0 \le x < 2, \int_x^6 k(t)\,dt \le \pi - 4 < 0 (at most the quarter disk, then the two triangles, -4), so B(x) > 0.
People pour in fastest at 4 pm, when r is largest, but P is then only 900 and still climbing, since r > 0. The count grows as long as r > 0, so it peaks when r crosses from positive to negative: 1{,}700 people at 6 pm.
By additivity, \int_c^x f(t)\,dt - \int_a^x f(t)\,dt = \int_c^a f(t)\,dt, a number with no x in it. Two functions that differ by a constant have the same largest-value point, since adding a constant moves every value by the same amount and keeps their order.
g' = f > 0 between consecutive zeros, so the continuous g is increasing on each piece, hence on [a, b]: g(a) < g(x) < g(b) for a < x < b.