The Second Fundamental Theorem of Calculus - At A Glance

As if one Fundamental Theorem of Calculus wasn't enough, there's a second one. The first FTC says how to evaluate the definite integral

if you know an antiderivative of f. The second FTC says how to build an antiderivative of f if you don't know one already.

Given a continuous function f and a constant a, we can define a new function F(x) by integrating f from a to x:

We can think of this new function F as an accumulation function—it accumulates the weighted area under f from a to x.

Given a continuous function f, we can build a new function F(x) by picking a lower limit of integration a and integrating f from a to x.

The Second Fundamental Theorem of Calculus says that when we build a function this way, we get an antiderivative of f.

Second Fundamental Theorem of Calculus:

Assume f(x) is a continuous function on the interval I and a is a constant in I. Define a new function F(x) by

Then F(x) is an antiderivative of f(x)—that is, F '(x) = f(x) for all x in I.

That business about the interval I is to make sure we only get limits of integration that are are reasonable for your function. Some things that wouldn't be reasonable:

Choosing a = -1 for the function
. Then

doesn't make sense because if x ≠ -1 we're trying to integrate a function over an interval where the function isn't even entirely defined.
Choosing a = 1 for the function but then trying to evaluate

Again, we're trying to integrate a function over an interval where the function isn't entirely defined.
Choosing a = 0 for the function . How would you evaluate

The problem is that a = 0 isn't in an interval on which f is continuous. Actually, we can do stuff with this integral—it just requires some extra work, and we haven't gotten there yet.
Choosing a = -1 for the function and trying to evaluate

If I is any interval that contains both -1 and 1, the function isn't continuous on that interval. Again, we can do stuff with this integral, but we haven't learned how yet.

We can condense this discussion into a short warning.

Be Careful: Whenever you go to evaluate an integral, make sure the limits of integration are reasonable for the function you're integrating.

If a function is discontinuous at or between the limits of integration we may still be able to evaluate the integral, but it requires a little extra sneakiness that we haven't gotten to yet.

So far, we've used the second FTC to build antiderivatives for things we already have derivatives for. That's not very exciting. Why go to all the trouble of writing

to get an antiderivative for sin x, when we already know that

-cos x

is an antiderivative of sin x?

It turns out there are many functions whose antiderivatives we don't know how to find. Some we'll learn how to find using various techniques of integration. However, there are many functions whose antiderivatives can't be written down with a nice formula. We can't write down a nice formula for the antiderivative of

e^{e^x}

for example - not just because we don't know how, but because it can't be done.

The second FTC lets us build an antiderivative of f(x) even in the cases where it's impossible to write a nice formula for an antiderivative.

Examples
Exercises

Example 1

Let f(x) = sin x and a = 0. Define

Find

(a) F(π)

(b)

(c)

Show Next Step

Example 2

Let . Find each value and represent each value using a graph of the function 2t.

(a) F(2)

(b) F(1)

(d) F(-1)

(e) F(-2)

Show Next Step

Exercise 1

Let f(x) = sin x and a = 0. Define

Find each value and represent using a graph of sin t.

(a) F(-π)

(b) F(2π)

(c)

(d) Consider all real values of x. For what values of x is F(x)

(i) positive?

(ii) negative?

(iii) zero?

Show Answer Gimme a Hint

Hint

Look at a graph of sin t, and what weighted area the integral of sin t from 0 to x accumulates.

Answer

(a)

Because we're integrating to the left, this value is the negative of the weighted area between sin t and the t-axis on [-π, 0]. So it's the positive value of this area:

(b)

This makes sense, because the weighted areas between sin t and the t-axis on [0, 2π] cancel out:

(c)

Since we're integrating to the left, this is the negative of the weighted area between sin t and the t-axis on :

The weighted area on is zero, because the areas above and below the axis on this interval are equal:

We're left with the negative of the negatively weighted area between sin t and the axis on , which is this area:

(d) We're integrating starting at 0. When we integrate sin t on the interval [0, 0] we get 0, so

When x is any integer multiple of 2π (positive or negative),

because the weighted areas the integral accumulates cancel each other out:

For any value of x in (0,π], the function

accumulates a positively weighted area, so the value of the function will be positive:

For any value of x in (π, 2π), the area below the t-axis is smaller than the area above the t-axis, so the function

is still positive:

Since , when we move x into the interval (2π, 3π] the values of F(x) are still positive:

And similarly for x in the interval (3π, 4π):

We can generalize to say that F(x) is zero when x is an integer multiple of 2π and positive for all positive values of x that aren't multiples of 2π.

When x is a negative multiple of 2π, we know that F(x) = 0. What about when x < 0 is not a multiple of 2π?

When x < 0 the function F(x) integrates to the left, so we count areas below the axis positively and areas above the axis negatively.

For x in [-π, 0) the function F(x) counts this area positively. This means F(x) > 0 for x in [-π, 0).

For x in (-2π, -π) the function F(x) considers the area below the axis positively, and the area above the axis negatively. Since there's more area below the axis than above the axis, F(x) is positive.

Generalizing, F(x) will be positive for all x < 0 that aren't multiples of 2π.

Putting all our observations together, we can finally answer the question.

(i) F(x) is positive for all x that aren't multiples of 2π.

(ii) F(x) is never negative.

(iii) F(x) is zero for all x that are multiples of 2π.

In the previous integral we used 0 as the lower limit of integration, but we can build a similar function using any lower limit of integration we want.