Free AP Calculus AB Integration and Accumulation of Change

AP Calculus AB · Integration and Accumulation of Change · free lesson

The Fundamental Theorem and Accumulation Functions

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

A function made by an area

So far a definite integral $\int_a^b f(t)\,dt$ has been one number: the signed area between the graph of $f$ and the axis from $a$ to $b$ (area above the axis counts positive, area below counts negative). Now keep the left end fixed and let the right end move. Each value of the right end $x$ gives a number, so the integral becomes a function of $x$ :

Definition.

An accumulation function is a function defined by an integral with a variable upper limit:

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

$g(x)$ is the signed area under the graph of $f$ from $t = a$ to $t = x$ . The letter $t$ is only the variable of integration; the input of $g$ is the upper limit $x$ . At the starting point, $g(a) = \int_a^a f(t)\,dt = 0$ : no width, no area.

Take the function $f$ whose graph is drawn below, on $0 \le t \le 8$ . It is made of three line segments and a semicircle: from $(0, 0)$ up to $(2, 2)$ , level at height 2 to $(3, 2)$ , down to $(4, 0)$ , and then a semicircle of radius 2 with center $(6, 0)$ below the axis. Let $g(x) = \int_0^x f(t)\,dt$ . Watch the shaded area grow as $x$ moves from 0 to 8: blue area adds to $g$ , red area takes away.

Each value of $g$ is an area of a simple shape. Find them in order, adding piece by piece:

$x$	New piece from the last $x$	Its signed area	$g(x)$
0	none	0	0
2	triangle, base 2, height 2	$2$	$2$
3	rectangle, width 1, height 2	$2$	$4$
4	triangle, base 1, height 2	$1$	$5$
6	quarter circle of radius 2, below the axis	$-{1}\pi$	${5} - {1}\pi \approx {1.858}$
8	another quarter circle, below the axis	$-{1}\pi$	${5} - {2}\pi \approx {-1.283}$

The triangle's area is $\frac{1}{2}(\text{base})(\text{height})$ , and a quarter of a circle of radius 2 has area $\frac{1}{4}\pi (2)^2 = {1}\pi$ . Notice what the numbers do: $g$ grows while $f$ is above the axis, is largest at $x = 4$ , where the graph of $f$ crosses down, and shrinks after it. The graph of $g$ (blue) draws itself next to $f$ (red):

The Fundamental Theorem of Calculus

How fast does $g$ grow? Move $x$ a little to the right, by a small width $h$ . The area gains a thin strip that is almost a rectangle of width $h$ and height $f(x)$ . So $g(x + h) - g(x) \approx f(x)\,h$ , and the rate of growth $\frac{g(x + h) - g(x)}{h}$ is close to $f(x)$ . As $h \to 0$ the strip becomes exactly such a rectangle, and the rate becomes $f(x)$ .

Drag the width $h$ down toward 0: the strip's area divided by its width, $\frac{g(x + h) - g(x)}{h}$ , closes in on $f(x)$ , wherever $x$ is. That is the theorem.

Rule.

The Fundamental Theorem of Calculus (accumulation form). If $f$ is continuous on an interval containing $a$ , then $g(x) = \int_a^x f(t)\,dt$ is differentiable there, and

\begin{aligned} g'(x) &= \frac{d}{dx}\int_a^x f(t)\,dt = f(x) \end{aligned}

The derivative of the accumulated area is the height of the graph at the moving end. Then $g''(x) = f'(x)$ wherever $f$ is differentiable.

For the $g$ above: $g'(1) = f(1) = {1}$ , and $g'(5) = f(5) = {-\sqrt{3}}$ , a negative rate, since at $x = 5$ the moving end is adding area below the axis. $g''(x) = f'(x)$ is the slope of the graph of $f$ : on $3 < x < 4$ the segment falls 2 for each 1 across, so $g''(3.5) = -2$ . Read the three letters this way: $g$ is an area, $g'$ is a height of $f$ , $g''$ is a slope of $f$ .

The theorem needs no formula for the area. For example, $\cos(t^2)$ has no antiderivative made of the usual functions, yet $\frac{d}{dx}\int_1^x \cos(t^2)\,dt = \cos(x^2)$ at once. The lower limit $a$ does not appear in the answer: moving it adds a constant to $g$ , and a constant has derivative 0.

When the upper limit is a function of x

If the upper limit is $u(x)$ instead of $x$ , then $\int_a^{u(x)} f(t)\,dt$ is $g(u(x))$ , a composite function. By the chain rule its derivative is $g'(u(x))\,u'(x) = f(u(x))\,u'(x)$ : put the upper limit into $f$ , then multiply by the derivative of the upper limit.

Rule.

\begin{aligned} \frac{d}{dx}\int_a^{u(x)} f(t)\,dt &= f(u(x))\,u'(x) \end{aligned}

For $H(x) = \int_0^{x^3} \sqrt{1 + t^2}\,dt$ , here $f(t) = \sqrt{1 + t^{2}}$ and $u(x) = x^{3}$ , with $u'(x) = {3x^{2}}$ . So

\begin{aligned} H'(x) &= f(x^3) \cdot {3x^{2}} = {3\sqrt{1 + x^{6}}x^{2}} \end{aligned}

and $H'(1) = {3\sqrt{2}}$ . When both limits move, split at a fixed number $a$ : $\int_{v(x)}^{u(x)} f(t)\,dt = \int_a^{u(x)} f(t)\,dt - \int_a^{v(x)} f(t)\,dt$ , so its derivative is $f(u(x))\,u'(x) - f(v(x))\,v'(x)$ .

Note.

On the AP exam the usual slips are: leaving out the factor $u'(x)$ of the chain rule; writing $f(u(x)) - f(a)$ , as if the lower limit were put in too; integrating $f$ when the question asks for the derivative; and taking $g''(c)$ to be $f''(c)$ , when it is $f'(c)$ . In a free response, state the theorem in one line: " $g'(x) = f(x)$ by the Fundamental Theorem of Calculus", then read the value off the graph.

Worked examples

Each example is one kind of question the AP exam asks about accumulation functions. Two facts do all the work: $g(x) = \int_a^x f(t)\,dt$ is a signed area, and $g'(x) = f(x)$ by the Fundamental Theorem of Calculus (with the chain rule when the upper limit is a function of $x$ ). Every number in the working is found by the engine.

Example.

1. Values of g, g' and g'' from a graph. The graph of $f$ on $-2 \le t \le 6$ , shown below, is made of a semicircle and two line segments. The semicircle has center $(0, 0)$ and radius 2 and lies above the axis; the segments join $(2, 0)$ to $(4, -2)$ and $(4, -2)$ to $(6, 0)$ . Let $g(x) = \int_{-2}^x f(t)\,dt$ . Find $g(2)$ , $g(5)$ , $g'(1)$ and $g''(5)$ .

$g(2)$ is the area from $-2$ to 2: the whole semicircle, $\frac{1}{2}\pi(2)^2 = {2}\pi$ . So $g(2) = {2}\pi \approx {6.283}$ .

$g(5)$ adds the area from 2 to 5, which is below the axis. From 2 to 4 it is a triangle of base 2 and height 2, area $2$ . From 4 to 5 the segment rises from $-2$ to $-1$ , so that piece is a trapezoid; easier, take the triangle from 4 to 6 (area $2$ ) and remove the small triangle from 5 to 6 (base 1, height 1, area $\frac{1}{2}$ ). The piece from 4 to 5 has area $\frac{3}{2}$ . Both pieces count negative:

\begin{aligned} g(5) &= {2}\pi - {2} - {\frac{3}{2}} = {2}\pi - {\frac{7}{2}} \end{aligned}

By the Fundamental Theorem, $g'(1) = f(1)$ , a height on the semicircle: $f(1) = \sqrt{4 - 1^2} = {\sqrt{3}}$ . And $g''(5) = f'(5)$ , the slope of the segment from $(4, -2)$ to $(6, 0)$ : $g''(5) = {1}$ .

Result.

$g(2) = {2}\pi$ , $g(5) = {2}\pi - {\frac{7}{2}}$ , $g'(1) = {\sqrt{3}}$ , $g''(5) = {1}$ .

Example.

2. A derivative with no antiderivative needed. Let $F(x) = \int_3^x \frac{t^2}{t^3 + 1}\,dt$ . Find $F'(x)$ and $F'(2)$ .

The integrand is continuous for $t > -1$ , an interval that contains 3 and 2. By the Fundamental Theorem, put $x$ in for $t$ :

\begin{aligned} F'(x) &= \frac{x^2}{x^3 + 1} \end{aligned}

so $F'(2) = {\frac{4}{9}}$ . The lower limit 3 plays no part. Integrating first and then differentiating would give the same answer, but it is longer and here needs a logarithm you do not need.

Result.

$F'(x) = {\frac{x^{2}}{x^{3} + 1}}$ and $F'(2) = {\frac{4}{9}}$ .

Example.

3. The chain rule for an upper limit. Let $H(x) = \int_1^{x^2} \cos(\pi t)\,dt$ . Find $H'(x)$ and $H'(3)$ .

The upper limit is $u(x) = x^{2}$ , with $u'(x) = {2x}$ . Put the upper limit into the integrand, then multiply by its derivative:

\begin{aligned} H'(x) &= \cos(\pi x^2) \cdot {2x} \end{aligned}

At $x = 3$ : $u(3) = 9$ and $\cos(9\pi) = {-1}$ , and $u'(3) = {6}$ . So

\begin{aligned} H'(3) &= ({-1})({6}) = {-6} \end{aligned}

Forgetting the factor $u'(x)$ gives $\cos(9\pi) = {-1}$ , the commonest wrong answer.

Result.

$H'(x) = 2x\cos(\pi x^2)$ and $H'(3) = {-6}$ .

Example.

4. Both limits variable. Let $K(x) = \int_{2x}^{x^2} \frac{1}{1 + t^2}\,dt$ . Find $K'(1)$ .

Split at 0: $K(x) = \int_0^{x^2} \frac{1}{1 + t^2}\,dt - \int_0^{2x} \frac{1}{1 + t^2}\,dt$ . Each part takes the chain rule: the upper limit $x^{2}$ has derivative $2x$ , the lower limit $2x$ has derivative 2.

\begin{aligned} K'(x) &= \frac{1}{1 + (x^2)^2} \cdot 2x - \frac{1}{1 + (2x)^2} \cdot 2 = {\frac{2x}{1 + x^{4}} - \frac{2}{1 + 4x^{2}}} \end{aligned}

At $x = 1$ : $K'(1) = {\frac{3}{5}}$ . The lower limit's term is subtracted: the area starts at $2x$ , so moving that end to the right takes area away.

Result.

$K'(1) = {\frac{3}{5}}$ .

Example.

5. The line tangent to an accumulation function. For the function $g$ of Example 1, write an equation for the line tangent to the graph of $g$ at $x = 4$ .

A tangent line needs a point and a slope. The point: $g(4)$ is the semicircle minus the triangle from 2 to 4, $g(4) = {2}\pi - {2} \approx {4.283}$ . The slope: $g'(4) = f(4) = {-2}$ , read from the graph.

\begin{aligned} y &= {2}\pi - {2} + ({-2})(x - 4) \end{aligned}

Do not swap the two: the height of $f$ is the slope of $g$ , and the area is the value of $g$ . In the picture, $g$ (blue) is drawn from its areas; the tangent line touches it at $x = 4$ , where $g$ is falling.

Result.

$y = {2}\pi - {2} - 2(x - 4)$ .

Example.

6. In context, with a calculator. Rain falls into a tank at the rate $r(t) = 4 + \sin\left(\frac{t^2}{3}\right)$ gallons per hour, $t$ hours after noon. At noon the tank holds 30 gallons, and no water leaves it. (a) Write an expression for $W(x)$ , the number of gallons in the tank at time $x$ . (b) Find $W'(2)$ and say what it means. (c) Find $W(3)$ .

(a) The water at time $x$ is the starting 30 gallons plus what has come in since noon:

\begin{aligned} W(x) &= 30 + \int_0^x r(t)\,dt \end{aligned}

(b) By the Fundamental Theorem, $W'(x) = r(x)$ (the constant 30 has derivative 0). So $W'(2) = r(2) = 4 + \sin\left(\frac{4}{3}\right) \approx {4.972}$ . At 2 p.m. the amount of water in the tank is increasing at about $4.972$ gallons per hour.

(c) This needs the calculator's integral: $W(3) = 30 + \int_0^3 r(t)\,dt$ , and the calculator gives $\int_0^3 r(t)\,dt \approx {13.545}$ , so $W(3) \approx {43.545}$ gallons. Write the integral down before the number: the setup is part of the answer.

Result.

$W(x) = 30 + \int_0^x r(t)\,dt$ ; $W'(2) \approx {4.972}$ gallons per hour; $W(3) \approx {43.545}$ gallons.

Practice

Answer AP-style questions on the fundamental theorem and accumulation functions. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise the fundamental theorem and accumulation functions with new numbers