Part I of the Fundamental Theorem says that we can use antiderivatives to compute definite integrals. Part II turns this relationship around: It tells us that we can use the definite integral to construct antiderivatives.
To state Part II, we introduce the area function of f with lower limit a:
A(x) is sometimes called the cumulative area function. In the definition of A(x), we use t as the variable of integration to avoid confusion with x, which is the upper limit of integration. In fact, t is a dummy variable and may be replaced by any other variable.
In essence, we turn the definite integral into a function by treating the upper limit x as a variable (Figure 1). Note that A(a) = 0 because .
In some cases we can find an explicit formula for A(x) [Figure 2].
Find a formula for the area function .
Solution The function is an antiderivative for f(t) = t2. By FTC I,
Note, in the previous example, that the derivative of A(x) is f(x) itself:
FTC II states that this relation always holds: The derivative of the area function is equal to the original function.
Assume that f(x) is continuous on an open interval I and let a ∈ I. Then the area function
is an antiderivative of f(x) on I; that is, A′ (x) = f(x). Equivalently,
Furthermore, A(x) satisfies the initial condition A(a) = 0.
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First, we use the additivity property of the definite integral to write the change in A(x) over [x, x + h] as an integral:
In other words, A(x + h) −A(x) is equal to the area of the thin sliver between the graph and the x-axis from x to x + h in Figure 3.
To simplify the rest of the proof, we assume that f(x) is increasing (see Exercise 50 for the general case). Then, if h > 0, this thin sliver lies between the two rectangles of heights f(x) and f(x + h) in Figure 4, and we have
In this proof,
Now divide by h to squeeze the difference quotient between f(x) and f(x + h):
We have because f(x) is continuous, and , so the Squeeze Theorem gives us
A similar argument shows that for h < 0,
Again, the Squeeze Theorem gives us
Equations (1) and (2) show that A′(x) exists and A′(x) = f(x).
Many applications (in the sciences, engineering, and statistics) involve functions for which there is no explicit formula. Often, however, these functions can be expressed as definite integrals (or as infinite series). This enables us to compute their values numerically and create plots using a computer algebra system. Figure 5 shows a computer-generated graph of an antiderivative of f(x) = sin(x2), for which there is no explicit formula.
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Let F(x) be the particular antiderivative of f(x) = sin (x2) satisfying . Express f(x) as an integral.
Solution According to FTC II, the area function with lower limit is an anti-derivative satisfying :
Find the derivative of
and calculate A′(2), A′(3), and A(2).
Solution By FTC II, . In particular,
On the other hand, .
The FTC shows that integration and differentiation are inverse operations. By FTC II, if you start with a continuous function f(x) and form the integral , then you get back the original function by differentiating:
On the other hand, by FTC I, if you differentiate first and then integrate, you also recover f(x) [but only up to a constant f(a)]:
When the upper limit of the integral is a function of x rather than x itself, we use FTC II together with the Chain Rule to differentiate the integral.
Find the derivative of
Solution FTC II does not apply directly because the upper limit is x2 rather than x. It is necessary to recognize that G(x) is a composite function with outer function :
FTC II tells us that A′(x) = sin x, so by the Chain Rule,
G′(x) = A′(x2) · (x2)′ = sin(x2) · (2x) = 2x sin(x2)
Alternatively, we may set u = x2 and use the Chain Rule as follows:
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FTC II tells us that A′(x) = f(x), or, in other words, f(x) is the rate of change of A(x). If we did not know this result, we might come to suspect it by comparing the graphs of A(x) and f(x). Consider the following:
These observations show that f(x) “behaves” like A′(x), as claimed by FTC II.