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The first two theorems of this section constitute the basic “techniques of integration” taught in a calculus course.However, the careful formulations of these standard methods of evaluating integrals have some subtle points, i.e., some hypotheses.Calculus students are rarely told about these details.

The first two theorems of this section constitute the basic “techniques of integration” taught in a calculus course.However, the careful formulations of these standard methods of evaluating integrals have some subtle points, i.e., some hypotheses.Calculus students are rarely told about these details.

Integration by parts formula

Let f and g be integrable functions on [ a , b ] , and as usual let F and G denote the functions defined by

F ( x ) = a x f , and G ( x ) = a x g .

Then

a b f G = [ F ( b ) G ( b ) - F ( a ) G ( a ) ] - a b F g .

Or, recalling that f = F ' and g = G ' ,

a b F ' G = [ F ( b ) G ( b ) - F ( a ) G ( a ) ] - a b F G ' .
  1. Prove the preceding theorem. HINT: Replace the upper limit b by a variable x , and differentiate both sides. By the way, how do we know that the functions F g and f G are integrable?
  2. Suppose f and g are integrable functions on [ a , b ] and that both f ' and g ' are continuous on ( a , b ) and integrable on [ a , b ] . (Of course f ' and g ' are not even defined at the endpoints a and b , but they can still be integrable on [ a , b ] . See the remark following [link] .) Prove that
    a b f g ' = [ f ( b ) g ( b ) - f ( a ) g ( a ) ] - a b f ' g .

Integration by substitution

Let f be a continuous function on [ a , b ] , and suppose g is a continuous, one-to-one function from [ c , d ] onto [ a , b ] such that g is continuously differentiable on ( c , d ) , and such that a = g ( c ) and b = g ( d ) . Assume finally that g ' is integrable on [ c , d ] . Then

a b f ( t ) d t = c d f ( g ( s ) ) g ' ( s ) d s .

It follows from our assumptions that the function f ( g ( s ) ) g ' ( s ) is continuous on ( a , b ) and integrable on [ c , d ] . It also follows from our assumptions that g maps the open interval ( c , d ) onto the open interval ( a , b ) . As usual, let F denote the function on [ a , b ] defined by F ( x ) = a x f ( t ) d t . Then, by part (2) of the Fundamental Theorem, F is differentiable on ( a , b ) , and F ' = f . Then, by the chain rule, F g is continuous and differentiable on ( c , d ) and

( F g ) ' ( s ) = F ' ( g ( s ) ) g ' ( s ) = f ( g ( s ) ) g ' ( s ) .

So, by part (3) of the Fundamental Theorem, we have that

c d f ( g ( s ) ) g ' ( s ) d s = c d ( F g ) ' ( s ) d s = ( F g ) ( d ) - ( F g ) ( c ) = F ( g ( d ) ) - F ( g ( c ) ) = F ( b ) - F ( a ) = a b f ( t ) d t ,

which finishes the proof.

  1. Prove the “Mean Value Theorem” for integrals: If f is continuous on [ a , b ] , then there exists a c ( a , b ) such that
    a b f ( t ) d t = f ( c ) ( b - a ) .
  2. (Uniform limits of differentiable functions. Compare with [link] .) Suppose { f n } is a sequence of continuous functions on a closed interval [ a , b ] that converges pointwise to a function f . Suppose that each derivative f n ' is continuous on the open interval ( a , b ) , is integrable on the closed interval [ a , b ] , and that the sequence { f n ' } converges uniformly to a function g on ( a , b ) . Prove that f is differentiable on ( a , b ) , and f ' = g . HINT: Let x be in ( a , b ) , and let c be in the interval ( a , x ) . Justify the following equalities, and use them together with the Fundamental Theorem to make the proof.
    f ( x ) - f ( c ) = lim ( f n ( x ) - f n ( c ) ) = lim c x f n ' = c x g .

We revisit now the Remainder Theorem of Taylor, which we first presented in [link] . The point is that there is another form of this theorem, the integral form,and this version is more powerful in some instances than the original one, e.g., in the general Binomial Theorem below.

Integral form of taylor's remainder theorem

Let c be a real number, and let f have n + 1 derivatives on ( c - r , c + r ) , and suppose that f ( n + 1 ) I ( [ c - r , c + r ] ) . Then for each c < x < c + r ,

f ( x ) - T ( f , c ) n ( x ) = c x f ( n + 1 ) ( t ) ( x - t ) n n ! d t ,

where T f n denotes the n th Taylor polynomial for f .

Similarly, for c - r < x < c ,

f ( x ) - T ( f , c ) n ( x ) = x c f ( n + 1 ) ( t ) ( x - t ) n n ! d t .

Prove the preceding theorem.

HINT: Argue by induction on n , and integrate by parts.

REMARK We return now to the general Binomial Theorem, first studied in [link] . The proof given there used the derivative form of Taylor's remainder Theorem,but we were only able to prove the Binomial Theorem for | t | < 1 / 2 . The theorem below uses the integral form of Taylor's Remainder Theorem in its proof, and it gives the full binomial theorem, i.e., for all t for which | t | < 1 .

General binomial theorem

Let α = a + b i be a fixed complex number. Then

( 1 + t ) α = k = 0 α k t k

for all t ( - 1 , 1 ) .

For clarity, we repeat some of the proof of [link] . Given a general α = a + b i , consider the function g : ( - 1 , 1 ) C defined by g ( t ) = ( 1 + t ) α . Observe that the n th derivative of g is given by

g ( n ) ( t ) = α ( α - 1 ) ... ( α - n + 1 ) ( 1 + t ) n - α .

Then g C ( ( - 1 , 1 ) ) .

For each nonnegative integer k define

a k = g ( k ) ( 0 ) / k ! = α ( α - 1 ) ... ( α - k + 1 ) k ! = α k ,

and set h ( t ) = k = 0 a k t k . The radius of convergence for the power series function h is 1, as was shown in [link] . We wish to show that g ( t ) = h ( t ) for all - 1 < t < 1 . That is, we wish to show that g is a Taylor series function around 0. It will suffice to show that the sequence { S n } of partial sums of the power series function h converges to the function g . We note also that the n th partial sum is just the n th Taylor polynomial T g n for g .

Now, fix a t strictly between 0 and 1 . The argument for t 's between - 1 and 0 is completely analogous.. Choose an ϵ > 0 for which β = ( 1 + ϵ ) t < 1 . We let C ϵ be a numbers such that | α n | C ϵ ( 1 + ϵ ) n for all nonnegative integers n . See [link] . We will also need the following estimate, which can beeasily deduced as a calculus exercise (See part (d) of [link] .). For all s between 0 and t , we have ( t - s ) / ( 1 + s ) t . Note also that, for any s ( 0 , t ) , we have | ( 1 + s ) α | = ( 1 + s ) a , and this is trapped between 1 and ( 1 + t ) a . Hence, there exists a number M t such that | ( 1 + s ) α - 1 | M t for all s ( - 0 , t ) . We will need this estimate in the calculation that follows.

Then, by the integral form of Taylor's Remainder Theorem, we have:

| g ( t ) - k = 0 n a k t k | = | g ( t ) - T g n ( t ) | = | 0 t g ( n + 1 ) ( s ) ( t - s ) n n ! d s | = | 0 t ( n + 1 ) × α n + 1 ( 1 + s ) α - n - 1 ( t - s ) n d s | 0 t | α n + 1 | | ( 1 + s ) α - 1 | ( n + 1 ) | ( t - s 1 + s | n d s 0 t | α n + 1 | M t ( n + 1 ) t n d s C ϵ M t ( n + 1 ) 0 t ( 1 + ϵ ) n + 1 t n d s = C ϵ M t ( n + 1 ) ( 1 + ϵ ) n + 1 t n + 1 = C ϵ M t ( n + 1 ) β n + 1 ,

which tends to 0 as n goes to , because β < 1 . This completes the proof for 0 < t < 1 .

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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ramon Reply
Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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Asali
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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China
Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Yasmin
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Cesar
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Ramkumar Reply
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AMJAD
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Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
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Azam
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
bamidele Reply
how did you get the value of 2000N.What calculations are needed to arrive at it
Smarajit Reply
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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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