Accumulation Functions

The upper limit is the variable

x

, so the variable of integration needs another letter. It is a dummy variable, and

t

is the usual choice.

When f ≥ 0 and x > a, g(x) is the area of the region from a to x.

For

x > a

, where the graph of

f

is below the axis, that part of the region counts as negative, as in any integral. At

x = a

the interval has no width, so

g(a) = 0

. For

x < a

the limits are reversed, and

g(x) = \int_{a}^{x} f(t)\,dt = -\int_{x}^{a} f(t)\,dt.

Interactive: The Accumulation Function

In the graph below,

f(t) = t^2

a = 0

, and the accumulation function is named

F

instead of

g

. Drag

x

to the left of

0

as well. There

t^2

is still positive, but

F(x)

is negative, because the limits are reversed.

Example 1

The graph of

f

[-2, 8]

consists of three line segments and a semicircle of radius

2

below the axis, as shown. Let

g(x) = \int_{0}^{x} f(t)\,dt.

Find

g(2)

g(4)

g(8)

, and

g(-2)

The graph of f: segments on [−2, 4], and a semicircle below the axis on [4, 8].

From

0

2

the region is a rectangle of height

2

, so

g(2) = 4

. From

2

4

it adds a triangle of base

2

and height

2

, of area

2

, so

g(4) = 6

. The half-disk on

[4, 8]

lies below the axis and has area

\frac{1}{2}\pi(2)^2 = 2\pi

, so

g(8) = 6 - 2\pi \approx -0.28.

For

x = -2

the limits are reversed. The triangle on

[-2, 0]

has area

2

, so

g(-2) = -\int_{-2}^{0} f(t)\,dt = -2.

The Second Fundamental Theorem

When

x

increases by a small amount

h

g

gains the area of a thin strip from

x

x + h

. The strip is nearly a rectangle of height

f(x)

and width

h

, so

g(x + h) - g(x) \approx f(x)\,h

The strip from x to x + h.

The strip gives the reason. For

h > 0

, the change

g(x + h) - g(x)

is the integral of

f

from

x

x + h

, and by the Mean Value Theorem for Integrals it equals

f(c)\,h

for some

c

between

x

and

x + h

. For

h < 0

the strip is to the left of

x

, but the same theorem on

[x + h, x]

gives the same equation. So the difference quotient

\frac{g(x + h) - g(x)}{h}

equals

f(c)

. As

h \to 0

c

is squeezed to

x

, and

f(c) \to f(x)

because

f

is continuous.

The College Board calls this theorem and the one that evaluates an integral as

F(b) - F(a)

together the Fundamental Theorem of Calculus.

Example 3

Find each derivative.

\begin{aligned} &\text{(a)}\;\; \frac{d}{dx}\int_{\pi}^{x} \sin t\,dt \\[10pt] &\text{(b)}\;\; \frac{d}{dx}\int_{x}^{5} \cos(t^2)\,dt \end{aligned}

(a) By the theorem, the derivative is

\sin x

. The lower limit

\pi

only adds a constant to the integral, and a constant has derivative

0

(b) Here

x

is the lower limit. Reverse the limits first:

\begin{aligned} &\frac{d}{dx}\int_{x}^{5} \cos(t^2)\,dt \\[4pt] &\qquad = \frac{d}{dx}\left(-\int_{5}^{x} \cos(t^2)\,dt\right) \\[4pt] &\qquad = -\cos(x^2). \end{aligned}

A function as a limit of integration

With

g(x) = \int_a^x f(t)\,dt

, an integral whose upper limit is a function

u(x)

g(u(x))

, a composition. The chain rule gives

g'(u(x))\,u'(x)

, and

g' = f

When both limits are functions

u

and

v

, differentiable with values in

I

, split the integral at

a

\begin{aligned} &\int_{v(x)}^{u(x)} f(t)\,dt \\[4pt] &\qquad = \int_{a}^{u(x)} f(t)\,dt - \int_{a}^{v(x)} f(t)\,dt. \end{aligned}

Each integral on the right is one of the kind above, so

\begin{aligned} &\frac{d}{dx}\int_{v(x)}^{u(x)} f(t)\,dt \\[4pt] &\qquad = f\bigl(u(x)\bigr)\,u'(x) - f\bigl(v(x)\bigr)\,v'(x). \end{aligned}

Example 4

Find each derivative. In (b),

x > 0

. In (c),

f

is the function graphed in Example 1; evaluate the derivative at

x = 2.5

\begin{aligned} &\text{(a)}\;\; \frac{d}{dx}\int_{0}^{\sin x} e^{t^2}\,dt \\[10pt] &\text{(b)}\;\; \frac{d}{dx}\int_{x}^{3x} \frac{1}{t}\,dt \\[10pt] &\text{(c)}\;\; \frac{d}{dx}\int_{0}^{2x} f(t)\,dt \end{aligned}

(a) Here

u(x) = \sin x

and

u'(x) = \cos x

, so the derivative is

e^{\sin^2 x}\cos x.

(b) The upper limit is

3x

, with derivative

3

, and the lower limit is

x

, with derivative

1

\frac{1}{3x}\cdot 3 - \frac{1}{x}\cdot 1 = \frac{1}{x} - \frac{1}{x} = 0.

That agrees with the integral itself, which is

\ln(3x) - \ln x = \ln 3

for every

x > 0

, a constant.

(c) The derivative is

f(2x) \cdot 2

. At

x = 2.5

it is

2f(5)

. The point

t = 5

is on the semicircle, where

f(t) = -\sqrt{4 - (t - 6)^2}

, so

f(5) = -\sqrt{3}

and the derivative is

-2\sqrt{3} \approx -3.46

An antiderivative with a given value

By the Second Fundamental Theorem, an accumulation function of

f

is an antiderivative of

f

. So every function continuous on an interval has an antiderivative there. Now let

f

be continuous on an interval

I

containing

a

, and

F' = f

I

. Since

F

is an antiderivative of

f

F(x) - F(a)

is the integral of

f

from

a

x

, so for

x

I

F(x) = F(a) + \int_{a}^{x} f(t)\,dt.

Example 6

Solar panels charge a school's battery at a rate of

r(t) = 12t - 3t^2

kilowatt-hours per hour,

t

hours after charging begins, for

0 \le t \le 4

. At

t = 0

the battery holds

20

kilowatt-hours. Let

E(x)

be the energy stored in the battery at time

x

, for

0 \le x \le 4

. Write

E(x)

with an integral, and find

E(2)

and

E'(2)

The battery holds what it started with plus what the panels have added since:

\begin{aligned} E(x) &= 20 + \int_{0}^{x} (12t - 3t^2)\,dt \\[4pt] &= 20 + 6x^2 - x^3. \end{aligned}

The energy added in the first 2 hours is the area under r from 0 to 2.

E(2) = 20 + 24 - 8 = 36

kilowatt-hours: the

20

it started with and

16

added. By the Second Fundamental Theorem,

E'(2) = r(2) = 12

kilowatt-hours per hour, the rate the panels charge the battery at that moment.