Applying The Fundamental Theorem of Calculus The FTC is going to get used all the time. In fact, it’s used so often that it will probably never explicitly be listed as a topic once we’re through with this set of notes. Here’s the FTC in all its glory: The Fundamental Theorem of Calculus
∫ f ′ ( x ) dx = f (b ) − f ( a ) b
a
Now let’s take a quick look at what this means by considering the graphs of f ( x ) and f ′ ( x ) .
f ′ ( x ) with
f ( x ) showing that f ( 8 ) − f ( 0 ) = 14.76
∫ f ′ ( x ) dx = 14.76 8
0
Problem: Put this concept into your own words. What does the definite integral of a function’s derivative represent in relation to the original function?
Problem: Given f ′ ( x ) below, evaluate below.
∫
3
−1
f ′ ( x ) dx . Indicate what you just found on the graph of f ( x )
Graph of f ′ ( x )
Notice that the FTC is essentially an equation:
Graph of f ( x )
∫ f ′ ( x ) dx = f (b ) − f ( a ) . b
a
You can use it in any way you
can use an ordinary equation. Add things to both sides, divide both sides by something, etc. In the previous problem you might have noticed that it was actually much easier to find f ( 3) − f ( −1) than to find
∫
3
−1
f ′ ( x ) dx . It happens—and we need to get good at recognizing when one is better than the other.
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Problem: Let’s evaluate each of the following definite integrals using the FTC with proper notation throughout. Verify your answer using geometry (if possible). a.
∫
3
0
x dx
In this case you were probably better off doing the geometry.
b.
∫
4
−1
x dx
Notice how the FTC just “takes care of” the whole positive/negative region thing that you have to worry about when doing the geometry? c.
∫ ( 2x − 1) dx 3
1
1
d.
dx 1+ x 2 3 3
∫
In this case, your only hope was really to use the FTC. You can still try to shade the region though, so go ahead!
Problem: Use the FTC to evaluate each of the following definite integrals. 4⎛ 1 ⎞ 3 b. ∫ ⎜ 3 ⎟ dx a. ∫ ( 3t + 2 ) dt c. 2 ⎝ x ⎠ −1
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∫
π
0
sin ( x ) dx
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So, what about the properties that you already figured out by doing lots of definite integrals geometrically? Yes! They all still apply—and you should use them as often as possible. Problem: Evaluate each of the following definite integrals. a.
c.
∫ ( x + 1)( x − 1) dx 0
b.
6
∫
1
3
d.
32 x dx
1
⎛ 1⎞
∫ ⎜⎝ x ⎟⎠ dx 2
∫
x2 x1
f ′ ( x ) dx
That last one was really just the FTC…you just can’t write that often enough! Whenever you can make use of even and odd functions, you should do that. Here’s a set of problems where you have a chance to do that—do it whenever possible or you’re wasting time. Problem: Evaluate each of the given definite integrals. Be smart about properties! a.
∫
π 2
0
sin ( x ) dx
b.
∫
0
−π 2
sin ( x ) dx
π 2
c.
∫
sin ( x ) dx
−π 2
d.
∫
π 2
0
cos ( x ) dx
e.
∫
0
−π 2
cos ( x ) dx
π 2
f.
∫
cos ( x ) dx
−π 2
5
g.
∫
j.
∫ (x
0
2
1
x 3 dx
2
+ 5 ) dx
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h.
∫
k.
∫ (x
−5
−1
−2
x 3 dx
2
+ 5 ) dx
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i.
∫
23
l.
∫
π 4
−23
x 3 dx
−π 4
tan ( x ) dx
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There are some fun, puzzle-like problems where you’re just given a bunch of definite integrals and their values and asked to use them to figure out some other, related definite integral. Here are a couple to try. Problem: If
∫ f ( x ) dx = 8 and ∫ f ( x ) dx = −3 , then what is ∫ f ( x ) dx ? 4
12
12
−6
−6
4
Problem: Assuming that f ( x ) is an even, continuous function such that
∫ f ( x ) dx = 5 and 2
1
∫ f ( x ) dx = 8 , evaluate ∫ f ( x ) dx . 1
1
−2
0
Problem: Rewrite the sum
∫ f ( x ) dx + ∫ f ( x ) dx + ∫ f ( x ) dx as a single definite integral. 8
0
6
0
−2
8
Problem: Use the table of selected values of the differentiable function f ( x ) to answer the questions that follow. 5 10 12 16 x f ( x) -4 3 6 8 f ′( x) 4 1 5 3 a. Approximate
∫ f ( x ) dx 16
5
with a left Riemann sum and intervals indicated in the table.
b. Approximate f ′ (11) .
c. Evaluate
∫
16
5
f ′ ( x ) dx .
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You’ll be using u-substitution all the time when dealing with the FTC—like almost every time! There are two different techniques you’re going to bounce back and forth between. • Consider the related indefinite integral. Evaluate it with u-substitution. Revert back to the original variables. Evaluate with the FTC. • Make a choice of u, change everything including the bounds, and then just do the new integral with the FTC. Problem: Use u-substitution to rewrite the definite integral entirely in terms of u—including the bounds! Evaluate. π 4 π 3 4 dx a. ∫ b. ∫ sec 2 ( 5x ) dx c. ∫ sin 3 ( 5x ) cos ( 5x ) dx 1 π 3 π 6 1+ 2x
Problem: Evaluate each of the definite integrals below. a.
∫
5
1
3x + 2 dx
b.
∫
0
π 8
x cos ( 4x 2 ) dx
c.
∫
π 4
0
tan 2 ( 3x ) sec 2 ( 3x ) dx
Remember: Generally speaking, you want to make u be something in a radical, the argument of a trig function, a trig function itself (if it’s being raised to a power), or an entire exponential function. There can be several layers, and your ability to do the problem successfully will depend on how much you practice! Sometimes it’s not obvious at all what to use for a substitution and you’ll just be told. In those cases just go with it. Usually it’s foreshadowing something you’ll learn in a later Calculus course. Problem: Entirely rewrite the definite integral
∫
3
−1
dx 4 − x2
with the substitution x = 2sin (θ ) . Evaluate
if you feel like it.
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This page is really important and you really need to think about and understand what we do here. It will come up all the time and you can’t afford to waste time thinking about it when it does. There are three ways of looking at the FTC:
∫ f ′ ( x ) dx = f (b ) − f ( a ) f ( b ) = f ( a ) + ∫ f ′ ( x ) dx f ( a ) = f ( b ) − ∫ f ′ ( x ) dx b
The standard view:
a
b
To move forward:
a b
To move backwards:
a
Problem: Let’s use these rearrangements and a calculator. a. Find f ( 3) to three decimal points given that f (1) = 1.256 and f ′ ( x ) = cos 3x 2 .
( ) (
)
b. Find f ( 6 ) to three decimal points given that f (10 ) = 8.654 and f ′ ( x ) = ln x 2 − 4 .
⎛ π ex ⎞ c. If f ′ ( x ) = sin ⎜ and f ( 0 ) = 1 , then what is f ( 2 ) ? ⎝ 2 ⎟⎠
d. If the function f is defined by f ( x ) = x 3 + 2 and g ( x ) is an antiderivative of f such that g ( 3) = 5 , then what is the value of g (1) ? Remember: The FTC and a calculator will allow you to find values of a function even though you might not be able to find an antiderivative—in fact, an antiderivative might not even exist…that’s when Riemann sums are crazy useful. Problem: Given the graph of f ( x ) to the right and let g ( x ) = x + ∫ f ( t ) dt . Find the indicated values x
of g ( x ) . a. g ( −2 )
−2
b. g ( −6 ) c. g ( 2 )
d. g ( 4 )
e. g ( 5 )
f.
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g (6)
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Problems: Evaluate each of the following definite integrals. a.
c.
e.
∫
π 4
∫
3
0
− 3
∫
tan x sec 2 x dx
4x x2 + 1
4
1
(
dx
dy
2 y 1+ y
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)
2
3π
b.
∫
d.
∫ (1− cos ( 3t )) sin ( 3t ) dt
f.
∫
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2π
3cos 2 x sin x dx
π 3
π 6
1
−1
r 1− r 2 dr
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Okay, so, remember the Candidates Test? It becomes a big deal when mixed with the FTC. Here it is again, just in case: Candidates Test If f ( x ) is continuous on a closed interval then its absolute maximum and absolute minimum values occur at an end point of the domain or a critical point within the domain. Problem: f ( x ) is the function passing through the point ( 2,6 ) whose derivative, f ′ ( x ) , is shown below. Use this information to answer the questions.
a. Find and classify the critical points of f ( x ) on the interval −8 < x < 7 .
b. Find the x-coordinates of each point of inflection of f ( x ) . Justify.
c. Find the absolute maximum and absolute minimum values of f ( x ) on −8 ≤ x ≤ 7 .
d. Find any values c, −8 < c < 7 , such that the instantaneous rate of change at x = c equals the average rate of change of f ( x ) on −8 ≤ x ≤ 7 . If such a value exists, was it guaranteed to exist? Why or why not?
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Problem: Let f ( x ) = ∫ t t 2 + 1 dt . This is called an accumulation function. x
1
a. Use your calculator to fill in the table for the function f ( x ) . (Three decimals.) -2 -1 0 1 2 3 x
4
f ( x) b. Use your calculator to graph f ( x ) . c. Find a form of f ( x ) that does not have an integral in it by applying the FTC to f ( x ) . (This is a really uncommon thing to be able to do for accumulation functions, by the way.)
d. Find f ′ ( x ) and f ′′ ( x ) .
e. Let g ( x ) = ∫ t t 2 + 1 dt . Notice the lower bound is different? Graph f ( x ) and g ( x ) on the same x
0
calculator screen. What appears to be true about the graphs? A good way to check your intuition might be to graph their derivatives.
Accumulation functions are one of the most common type of functions you’ll find yourself dealing with because they really test how well you’re understanding the Calculus we’re learning. Calc AB Notes 16
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