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The Fundamental Theorem of Calculus and Mean Value Theorem 2

We’ve learned two different branches of calculus so far: differentiation and integration. Finding slopes of tangent lines and finding areas under curves seem unrelated, but in fact, they are very closely related. It was Isaac Newton’s teacher at Cambridge University, a man name Isaac Barrow (1630 – 1677), who discovered that these two processes are actually inverse operations of each other in much the same way division and multiplication are.

It was Newton and Leibniz who exploited this idea and developed the calculus into its current form.

The Theorem Barrow discovered that states this inverse relation between differentiation and integration is called

The Fundamental Theorem of Calculus.

FToC1 bridges the antiderivative concept with the area problem.

This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.

To evaluate an integral, take the antiderivatives and subtract.

It can be proved and the proof can be found elsewhere (WEB)

It is a “shortcut” rule for integration: an easier way (from Riemann sums and other methods) to calculate definite integrals.

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FToC 1∫a

b

f ( x ) dx=F (b )−F(a)

3

4

FToC 2 ( part 1 )F ( x )=∫a

x

f ( t ) dt upper boundaryis x

where f is a continuous function on[a ,b ], and xvaries between a and b. Notice that this integral equationis a function of x, which appears as the upper limit of integration. If f (t) happens to be positive, and welet x ϵ ¿ ,then we can define F (x) as the area under the curve from a to x .

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F (x)=∫3

x

( t2+t+1 )=( t 3

3+ t 2

2+t)|

3

x

= x3

3+ x2

2+ x−33

2

ddx ( x3

3+ x2

2+x−33

2 )=x2+x+1

FToC 2( part 2)F' (x )= ddx∫a

x

f (t )dt=f (x)

F ( x )=∫−π

x

cos t dt F' ( x )=?

Directly:

F ' ( x )=cos x

Or, longer way:

F ' ( x )= ddx ∫−π

x

cos t dt= ddx

¿¿¿

¿ ddx (sin x−sin (−π))=cos x

What if upper limit is g(x) not x itself?

Lower limit of integration is a constant.

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1 1+t

xd dtdx∫ 2

11 x

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Or upper limit is constant and lower limit is x?

FToC 2 ,most general form ddx [∫h (x)

g(x)

f ( t )dt=f [g (x)] ∙ g' ( x )−f [h ( x ) ] ∙ h '(x )]

Example 10:Evaluate the following using the FTOC2, then if feeling bored verify by doing in the Loooooooong way.

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Solution:

Avgvalue=∫0

12

[50+14 sin( πt12 )]dt12−0

= 112 [50 t−14 12

πcos ( πt12 )]|0

12

=50+2 14π

≈58.913

Geometric interpretation: For a positive function f (x), there exists at least one number a≤c≤bsuch that the rectangle with base [a ,b]and height f (c )has the same area as the region under the graph of f (x)from a to b.

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Solution:

f (c )=∫−1

2

[1+ x2 ] dx

2−(−1)=1

3 [ x+ x3

3 ]|−1

2

=13 [2+ 8

3−(−1)−(−1

3 )]=2

1 + x2 = 2 ⟹ c=±1 for x = ±1

For positive function f (x)=1+x2, there exist two numbers c=±1 such that the rectangle with base [−1 ,1]and height f (±1)has the same area as the region under the graph of f (x)from -1 to 2.

Solution:

The average rate of change of a car’s position over an interval is represented graphically as the slope of the secant line to the graph position/time over the interval.

vavg=x( t2)−x (t 1)

t 2−t 1

Average value of velocity over the same interval is:

vavg=1

t2−t1∫t 1

t 2

v (t )dt

proof : 1t2−t1

∫t 1

t 2

v (t)dt= 1t 2−t 1

∫t 1

t 2

x ' (t)dt= 1t 2−t 1

[ x (t 2)−x (t 1)] by FToC 1

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Solution:

f avg=∑ f (x)

7≈ 40.429

Area

area = 1160 average=1160/30 ≈ 38.667

trapezoidal approximation is more accurate that LRAM (basically arithmetic mean)

Sometimes we might have to solve an integral equation! Being able to simplify definite integrals withvariables in the interval of integration is important. Here are a couple of examples showing an importantapplication that is important.

Solution:

1b∫0

b

(2+6 x−3 x2 )dx=3

2+3b−b2=3→b2−3b+1=0→b=3±√52

Solution:

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Solution:

In the previous example, the first method relied heavily upon our ability to find the antiderivative of theintegrand. This is not always easy, possible, or prudent! Being able to express a particular value of aparticular solution to a derivative as a definite integral is of paramount importance, especially when wedon’t know how to find a general antiderivative. (calculator can do easily definite integrals – see problem 20)

Hard Facts To Refute:

(a ) f (x )=∫ (2x2−2 )dx=23x3−2x+C

f (0 )=5=C

f ( x )=23x3−2x+5 f (2 )=19

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(b ) f (x)=f (0)+∫0

x

f ' (x )dx f (2)=5+∫0

2

( 2x2−2 )dx=193

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A. Where you are at any given time is a function of 1) where you started and 2) where you’ve gone fromyour starting point (displacement).B. What you have at any given moment is a function of 1) what you started with plus 2) what you’veaccumulated since then.When you accumulate at a variable rate, you can to use the definite integral to find your net accumulation.Important Idea of Accumulation********************************(* means VERY IMPORTANT)

Example 19:

If f ' ( x )=4 sin2(2 x) and f (2 )=−2 find (a) and integral equation for f(x) (b) f(3) (c) f(-2)

Solution:

a¿ f (x )=f (2 )+∫2

x

4 sin2 (2 x )dx=f (2 )+2∫2

x

(1−cos 4 x )dx=¿¿

¿ f (2 )+2 x|2x −1

2sin 4 x|

2

x

¿−2+2x−4−sin 4 x+ 12

sin (8)

f (x)=2x−6−sin 4 x+ 12

sin (8)

b¿ f (3 )=f (2 )+∫2

3

4 sin2 (2 x )dx=f (2 )+2∫2

3

(1−cos4 x )dx=¿¿

¿−2+2x|23 −1

2sin 4 x|

2

3

=−2+6−4−12

sin12+ 12

sin(8)

f (3 )=12

sin 8−12

sin 12

b¿ f (−2 )=−2+∫2

−2

4 sin2 (2x )dx=−2+2∫2

−2

(1−cos4 x )dx=¿¿

¿−2+2x|2−2 −1

2sin 4 x|

2

−2

=−2−4−4−12

sin(−8)+ 12

sin (8)

cos2 x=cos2 x−sin 2 x

cos2 x=1−2 sin2 x→ sin2 x=1−cos2x2

cos2 x=1−2 sin2 x→cos2 x=1+cos2x2

What one has now = What one started with + What one has accumulated since one startedThis can be expressed mathematically as

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f (−2 )=sin (8 )−10

Solution: Solution:

(a) g(2)=−π g (x)=g (2)+∫2

x

g ' (x)dx=−π+∫2

x

esin x dx

(b) g(π )=−π+∫2

π

esin xdx ≈−1.169 (c) g(π )=−π+∫2

0

esin x dx≈−7.378

Solution:

W ( t)=W (0)+∫0

t

W ' (t)dt⟹W (14)=180+W (t)|014

W (14 )=180+10 sin 7 π4

≈172.930duringChristmastime ? whocooked ?