5.1. CONSTRUCTING ACCURATE GRAPHS OF ANTIDERIVATIVES
267
(d) Recall that the Total Change Theorem tells us that
f (1) − f (0) =
1
0
f
′ (x) dx.
What is the exact value of f (1)?
(e) Use the given information and similar reasoning to that in (d) to determine the
exact value of f (2), f (3), f (4), f (5), and f (6).
(f) Based on your responses to all of the preceding questions, sketch a complete
and accurate graph of y = f (x) on the axes provided, being sure to indicate the
behavior of f for x < 0 and x > 6.
⊲⊳
Constructing the graph of an antiderivative
Preview Activity 5.1 demonstrates that when we can find the exact area under a given graph
on any given interval, it is possible to construct an accurate graph of the given function’s
antiderivative: that is, we can find a representation of a function whose derivative is the
given one. While we have considered this question at different points throughout our study,
it is important to note here that we now can determine not only the overall shape of the
antiderivative, but also the actual height of the antiderivative at any point of interest.
Indeed, this is one key consequence of the Fundamental Theorem of Calculus: if we
know a function f and wish to know information about its antiderivative, F, provided that
we have some starting point a for which we know the value of F(a), we can determine
the value of F(b) via the definite integral. In particular, since F(b) − F(a) =
b
a
f (x) dx, it
follows that
F(b) = F(a) +
b
a
f (x) dx.
(5.1)
Moreover, in the discussion surrounding Figure 4.34, we made the observation that
differences in heights of a function correspond to net-signed areas bounded by its derivative.
Rephrasing this in terms of a given function f and its antiderivative F, we observe that on
an interval [a, b],
differences in heights on the antiderivative (such as F(b) − F(a)) correspond
to the net-signed area bounded by the original function on the interval [a, b]
(
b
a
f (x) dx).
For example, say that f (x) = x 2 and that we are interested in an antiderivative of f that
satisfies F(1) = 2. Thinking of a = 1 and b = 2 in Equation (5.1), it follows from the
267
(d) Recall that the Total Change Theorem tells us that
f (1) − f (0) =
1
0
f
′ (x) dx.
What is the exact value of f (1)?
(e) Use the given information and similar reasoning to that in (d) to determine the
exact value of f (2), f (3), f (4), f (5), and f (6).
(f) Based on your responses to all of the preceding questions, sketch a complete
and accurate graph of y = f (x) on the axes provided, being sure to indicate the
behavior of f for x < 0 and x > 6.
⊲⊳
Constructing the graph of an antiderivative
Preview Activity 5.1 demonstrates that when we can find the exact area under a given graph
on any given interval, it is possible to construct an accurate graph of the given function’s
antiderivative: that is, we can find a representation of a function whose derivative is the
given one. While we have considered this question at different points throughout our study,
it is important to note here that we now can determine not only the overall shape of the
antiderivative, but also the actual height of the antiderivative at any point of interest.
Indeed, this is one key consequence of the Fundamental Theorem of Calculus: if we
know a function f and wish to know information about its antiderivative, F, provided that
we have some starting point a for which we know the value of F(a), we can determine
the value of F(b) via the definite integral. In particular, since F(b) − F(a) =
b
a
f (x) dx, it
follows that
F(b) = F(a) +
b
a
f (x) dx.
(5.1)
Moreover, in the discussion surrounding Figure 4.34, we made the observation that
differences in heights of a function correspond to net-signed areas bounded by its derivative.
Rephrasing this in terms of a given function f and its antiderivative F, we observe that on
an interval [a, b],
differences in heights on the antiderivative (such as F(b) − F(a)) correspond
to the net-signed area bounded by the original function on the interval [a, b]
(
b
a
f (x) dx).
For example, say that f (x) = x 2 and that we are interested in an antiderivative of f that
satisfies F(1) = 2. Thinking of a = 1 and b = 2 in Equation (5.1), it follows from the