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5.1. CONSTRUCTING ACCURATE GRAPHS OF ANTIDERIVATIVES
derivative f ), but G will be shifted vertically away from the graph of F, as pictured in red
in Figure 5.3. Note that G(1) − G(0) =
1
0
f (x) dx = 0.5, just as F(1) − F(0) = 0.5,, but
since G(0) = 3, G(1) = G(0) + 0.5 = 3.5, whereas F(1) = F(0) + 0.5 = 1.5, since F(0) = 1.
In the same way, if we assigned a different initial value to the antiderivative, say H(0) = −1,
we would get still another antiderivative, as shown in magenta in Figure 5.3.
This example demonstrates an important fact that holds more generally:
If G and H are both antiderivatives of a function f , then the function G − H must be
constant.
To see why this result holds, observe that if G and H are both antiderivatives of f , then
G ′ = f and H ′ = f . Hence,
d
dx [G(x) − H(x)] = G ′ (x) − H ′ (x) = f (x) − f (x) = 0. Since
the only way a function can have derivative zero is by being a constant function, it follows
that the function G − H must be constant.
Further, we now see that if a function has a single antiderivative, it must have infinitely
many: we can add any constant of our choice to the antiderivative and get another
antiderivative. For this reason, we sometimes refer to the general antiderivative of a
function f . For example, if f (x) = x 2 , its general antiderivative is F(x) =
1
3 x 3 + C, where
we include the “+C” to indicate that F includes all of the possible antiderivatives of f .
To identify a particular antiderivative of f , we must be provided a single value of the
antiderivative F (this value is often called an initial condition). In the present example,
suppose that condition is F(2) = 3; substituting the value of 2 for x in F(x) =
1
3 x 3 + C, we
find that
3 =
1
3
(2)
3 + C,
and thus C = 3 −
8
3 =
1
3 . Therefore, the particular antiderivative in this case is F(x) =
1
3 x 3 +
1
3 .
Activity 5.2.
For each of the following functions, sketch an accurate graph of the antiderivative that
satisfies the given initial condition. In addition, sketch the graph of two additional
antiderivatives of the given function, and state the corresponding initial conditions that
each of them satisfy. If possible, find an algebraic formula for the antiderivative that
satisfies the initial condition.
(a) original function: g(x) = |x| − 1;
initial condition: G(−1) = 0;
interval for sketch: [−2, 2]
(b) original function: h(x) = sin(x);
initial condition: H(0) = 1;
interval for sketch: [0, 4π]
5.1. CONSTRUCTING ACCURATE GRAPHS OF ANTIDERIVATIVES
derivative f ), but G will be shifted vertically away from the graph of F, as pictured in red
in Figure 5.3. Note that G(1) − G(0) =
1
0
f (x) dx = 0.5, just as F(1) − F(0) = 0.5,, but
since G(0) = 3, G(1) = G(0) + 0.5 = 3.5, whereas F(1) = F(0) + 0.5 = 1.5, since F(0) = 1.
In the same way, if we assigned a different initial value to the antiderivative, say H(0) = −1,
we would get still another antiderivative, as shown in magenta in Figure 5.3.
This example demonstrates an important fact that holds more generally:
If G and H are both antiderivatives of a function f , then the function G − H must be
constant.
To see why this result holds, observe that if G and H are both antiderivatives of f , then
G ′ = f and H ′ = f . Hence,
d
dx [G(x) − H(x)] = G ′ (x) − H ′ (x) = f (x) − f (x) = 0. Since
the only way a function can have derivative zero is by being a constant function, it follows
that the function G − H must be constant.
Further, we now see that if a function has a single antiderivative, it must have infinitely
many: we can add any constant of our choice to the antiderivative and get another
antiderivative. For this reason, we sometimes refer to the general antiderivative of a
function f . For example, if f (x) = x 2 , its general antiderivative is F(x) =
1
3 x 3 + C, where
we include the “+C” to indicate that F includes all of the possible antiderivatives of f .
To identify a particular antiderivative of f , we must be provided a single value of the
antiderivative F (this value is often called an initial condition). In the present example,
suppose that condition is F(2) = 3; substituting the value of 2 for x in F(x) =
1
3 x 3 + C, we
find that
3 =
1
3
(2)
3 + C,
and thus C = 3 −
8
3 =
1
3 . Therefore, the particular antiderivative in this case is F(x) =
1
3 x 3 +
1
3 .
Activity 5.2.
For each of the following functions, sketch an accurate graph of the antiderivative that
satisfies the given initial condition. In addition, sketch the graph of two additional
antiderivatives of the given function, and state the corresponding initial conditions that
each of them satisfy. If possible, find an algebraic formula for the antiderivative that
satisfies the initial condition.
(a) original function: g(x) = |x| − 1;
initial condition: G(−1) = 0;
interval for sketch: [−2, 2]
(b) original function: h(x) = sin(x);
initial condition: H(0) = 1;
interval for sketch: [0, 4π]