Topic-2-The Fundamental Theorem of Calculus
Transcript of Topic-2-The Fundamental Theorem of Calculus
Dr. Gizem SEYHAN ΓZTEPE
Topic-2-The Fundamental Theorem of Calculus
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The Fundamental Theorem of Calculus is appropriately named because itestablishes a connection between the two branches of calculus: differential calculusand integral calculus.
Differential calculus arose from the tangent problem, whereas integral calculusarose from a seemingly unrelated problem, the area problem. Newtonβs teacher atCambridge, Isaac Barrow (1630β1677), discovered that these two problems areactually closely related. In fact, he realized that differentiation and integration areinverse processes.
The Fundamental Theorem of Calculus gives the precise inverse relationshipbetween the derivative and the integral. It was Newton and Leibniz who exploitedthis relationship and used it to develop calculus into a systematic mathematicalmethod.
In particular, they saw that the Fundamental Theorem enabled them to computeareas and integrals very easily without having to compute them as limits of sumsas we did in Sections 1.1 and 1.2.
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The first part of the Fundamental Theorem deals withfunctions defined by an equation of the form
where π is a continuous function on [π, π] and π₯ variesbetween π and π. Observe that π depends only on π₯,which appears as the variable upper limit in the integral.
If π₯ is a fixed number, then the integral π
π₯π π‘ ππ‘ is a
definite number. If we then let π₯ vary, the number
π
π₯π π‘ ππ‘ also varies and defines a function of π₯ denoted
by π π₯ .
π π₯ = ΰΆ±π
π₯
π π‘ ππ‘ (1)
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If π happens to be a positive function, thenπ(π₯) can be interpreted as the area underthe graph of π from π to π₯, where π₯ can varyfrom π to π. (Think of as the βarea so farβfunction; see Figure 1.)
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If π is continuous on [π, π], then the functionπ defined by
is continuous on [π, π], and differentiable on (π, π), and πβ² π₯ = π π₯ .
This theorem says that the derivative of a definite integralwith respect to its upper limit is the integrand evaluated atthe upper limit.
π π₯ = ΰΆ±π
π₯
π π‘ ππ‘ , π β€ π₯ β€ π
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Example-1
Find the derivative of the function π π₯ = 0
π₯1 + π‘2ππ‘.
Solution
Since π π‘ = 1 + π‘2 is continuous by using FTC1,
πβ² π₯ = 1 + π₯2
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Example-2
Although a formula of the form π π₯ = π
π₯π π‘ ππ‘ may seem
like a strange way of defining a function, books on physics,chemistry, and statistics are full of such functions.
For instance, the Fresnel function
π π₯ = ΰΆ±0
π₯
sin( ΰ΅ππ‘2
2)ππ‘
is named after the French physicist Augustin Fresnel (1788β1827), who is famous for his works in optics. This functionfirst appeared in Fresnelβs theory of the diffraction of light
waves, but more recently it has been applied to the design ofhighways.
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Part 1 of the Fundamental Theorem tells us how to differentiate the Fresnel function:
πβ² π₯ = sin( ΰ΅ππ₯2
2)
This means that we can apply all the methods of differential calculus to analyze π.
Figure 2 shows the graphs of π π₯ = sin( ΰ΅ππ₯2
2) and the Fresnel function π π₯ =
0
π₯sin( ΰ΅ππ‘2
2)ππ‘
Figure 3 shows a larger part of the graph of .
Figure-2 Figure-3
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In Section 1.2 we computed integrals from the
definition as a limit of Riemann sums and we
saw that this procedure is sometimes long and difficult.
The second part of the Fundamental Theorem of Calculus,which follows easily from the first part, provides us with amuch simpler method for the evaluation of integrals.
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If π is continuous on [π, π], then
where πΉ is any antiderivative of π, that is, a functionsuch that πΉβ² = π.
ΰΆ±π
π
π π₯ ππ₯ = πΉ π β πΉ(π)
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Part 2 of the Fundamental Theorem states that if we know
an antiderivative πΉ of π, then we can evaluate π
ππ π₯ ππ₯
simply by subtracting the values of πΉ at the endpoints of the
interval [π, π]. Itβs very surprising that π
ππ π₯ ππ₯, which was
defined by a complicated procedure involving all of the values of π(π₯) for π β€ π₯ β€ π, can be found by knowing the values of πΉ(π₯) at only two points, π and π.
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Example-3
Evaluate the integral 1
3ππ₯dx.
Solution
The function π π₯ = ππ₯ is continuous everywhere and we know that an antiderivative is πΉ(π₯) = ππ₯, so Part 2 of the Fundamental Theorem gives
ΰΆ±1
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ππ₯ ππ₯ = πΉ 3 β πΉ 1 = π3 β π
Notice that FTC2 says we can use any antiderivative πΉ of π. So we may as well use the simplest one, namely πΉ π₯ = ππ₯ instead of ππ₯ + 7 ππ ππ₯ + πΆ.
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Note: We often use the notation
ΰΈΰΆ±π
π
π π₯ ππ₯ = πΉ(π₯)
π
π
= πΉ π β πΉ(π)
where πΉβ² = π.
Example-4
Find the area under the parabola π¦ = π₯2 from 0 to 1.
OR
Evaluate the integral0
1π₯2ππ₯.
Example-5
Evaluate 3
6 ππ₯
π₯.
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Example-6
Find the area under the cosine curve from 0 to b , where 0 β€
π β€π
2.
Example-7
What is wrong with the following calculation?
ΰΆ±β1
3 1
π₯2ππ₯ = ΰΈ
π₯β1
β1β1
3
=β1
3β 1 =
β4
3
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The Fundamental Theorem of Calculus say that differentiation andintegration are inverse processes. Each undoes what the other does.
The Fundamental Theorem of Calculus is unquestionably the mostimportant theorem in calculus and, indeed, it ranks as one of thegreat accomplishments of the human mind. Before it wasdiscovered, from the time of Eudoxus and Archimedes to the time ofGalileo and Fermat, problems of finding areas, volumes, and lengthsof curves were so difficult that only a genius could meet thechallenge. But now, armed with the systematic method that Newtonand Leibniz fashioned out of the Fundamental Theorem, we will seein the chapters to come that these challenging problems areaccessible to all of us.
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Theorem (Symmetry Rule)
Let π be a continuous function defined on [βπ, π]
(i) If π is an odd function, then πβ
ππ π₯ ππ₯ = 0
(ii) If π is an even function, then πβ
ππ π₯ ππ₯ = 2 0
ππ π₯ ππ₯.
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If πβ² is continuous on [π, π] and π is continuous on therange of π’ = π(π₯), then
ΰΆ±π
π
π π π₯ πβ² π₯ ππ₯ = ΰΆ±π(π)
π(π)
π π’ ππ’
This rule says that when using a substitution in adefinite integral, we must put everything in terms ofthe new variable , not only and but also the limits ofintegration.
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Example-8
Evaluate the following integrals:
a) 0
42π₯ + 1 ππ₯
b) 0
π2
4sin π₯
π₯ππ₯
c) 0
ππ₯ cos π₯2ππ₯
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Solution
a) 0
42π₯ + 1 ππ₯
The geometric interpretation of Example(8.a) is shown in Figure 2. The substitutionπ’ = 2π₯ + 1stretches the interval [0,4] by afactor of 2 and translates it to the right byunit. The Substitution Rule shows that thetwo areas are equal.
More examples will be solved in the class.
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Example
Find the average value of the function π π₯ = 1 + π₯2on theinterval [β1,2].
Solution
With π = β1 and π = 2, we have
πππ£π =1
(2 β (β1))ΰΆ±
β1
2
1 + π₯2 ππ₯ = 2
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