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5.2. THE SECOND FUNDAMENTAL THEOREM OF CALCULUS
one another.
Taking a different approach, say we begin with a function f (t) and differentiate with
respect to t. What happens if we follow this by integrating the result from t = a to t = x?
That is, what can we say about the quantity
x
a
d
dt
[ f (t)] dt?
Here, we use the First FTC and note that f (t) is an antiderivative of
d
dt [ f (t)] . Applying
this result and evaluating the antiderivative function, we see that
x
a
d
dt
[ f (t)] dt = f (t)
x
a
= f (x) − f (a).
Thus, we see that if we apply the processes of first differentiating f and then integrating
the result from a to x, we return to the function f , minus the constant value f (a). So in
this situation, the two processes almost undo one another, up to the constant f (a).
The observations made in the preceding two paragraphs demonstrate that differentiating and integrating (where we integrate from a constant up to a variable) are almost
inverse processes. In one sense, this should not be surprising: integrating involves antidifferentiating, which reverses the process of differentiating. On the other hand, we see that
there is some subtlety involved, as integrating the derivative of a function does not quite
produce the function itself. This is connected to a key fact we observed in Section 5.1,
which is that any function has an entire family of antiderivatives, and any two of those
antiderivatives differ only by a constant.
Activity 5.6.
Evaluate each of the following derivatives and definite integrals. Clearly cite whether
you use the First or Second FTC in so doing.
(a)
d
dx
x
4
e
t 2 dt
(b)
x
−2
d
dt
t 4
1 + t 4
dt
(c)
d
dx
1
x
cos(t
3 ) dt
(d)
x
3
d
dt
ln(1 + t
2 )
dt
(e)
d
dx
x 3
4
sin(t
2 ) dt
5.2. THE SECOND FUNDAMENTAL THEOREM OF CALCULUS
one another.
Taking a different approach, say we begin with a function f (t) and differentiate with
respect to t. What happens if we follow this by integrating the result from t = a to t = x?
That is, what can we say about the quantity
x
a
d
dt
[ f (t)] dt?
Here, we use the First FTC and note that f (t) is an antiderivative of
d
dt [ f (t)] . Applying
this result and evaluating the antiderivative function, we see that
x
a
d
dt
[ f (t)] dt = f (t)
x
a
= f (x) − f (a).
Thus, we see that if we apply the processes of first differentiating f and then integrating
the result from a to x, we return to the function f , minus the constant value f (a). So in
this situation, the two processes almost undo one another, up to the constant f (a).
The observations made in the preceding two paragraphs demonstrate that differentiating and integrating (where we integrate from a constant up to a variable) are almost
inverse processes. In one sense, this should not be surprising: integrating involves antidifferentiating, which reverses the process of differentiating. On the other hand, we see that
there is some subtlety involved, as integrating the derivative of a function does not quite
produce the function itself. This is connected to a key fact we observed in Section 5.1,
which is that any function has an entire family of antiderivatives, and any two of those
antiderivatives differ only by a constant.
Activity 5.6.
Evaluate each of the following derivatives and definite integrals. Clearly cite whether
you use the First or Second FTC in so doing.
(a)
d
dx
x
4
e
t 2 dt
(b)
x
−2
d
dt
t 4
1 + t 4
dt
(c)
d
dx
1
x
cos(t
3 ) dt
(d)
x
3
d
dt
ln(1 + t
2 )
dt
(e)
d
dx
x 3
4
sin(t
2 ) dt