A mathematical description of motion motivated

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Integral Calculus
A mathematical description of motion
motivated the creation of Calculus. Problem of
Motion Given x(t) find v(t)
Differential Calculus. Given v(t) find x(t)
Integral Calculus.
Derivatives and integrals are operations on
functions. One is the inverse of the other. This
is the content of the Fundamental theorem of
Calculus.
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Integral calculus is mainly due to the
contributions from the following well known
mathematicians.(The photographs are worth
watching since these names will appear many times
in the courses to follow.)
Isaac Newton
Gottfried Leibniz
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James Gregory
Pierre de Fermat
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Joseph Fourier
Cauchy
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Bernhard Riemann
Henri Lebesgue
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Some motivations 1. Suspension
bridges
The road deck hangs on vertical cables suspended
from the main cables. Problem We have to find
the optimal shape of the main cable.
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Mathematical description (Model)
Solution This is the basic problem of integral
calculus and we solve the problem by integration.
y(x) ? y'(x) dx ? µx dx µ (x2/2) C. The
main cable has a parabolic shape.
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2. Reduction formulae are useful to compute the
following
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REDUCTION FORMULAE Reduction formula for ?
sinn x dx where n is a positive integer. Let In
? sinnx dx ? sinn-1 x.sin x. dx ? u v
dx (say) We know that ? uv dx u (? v dx) - ?
(? v dx ) u1 dx In sinn-1 x (-cos x) - ?
(-cos x) (n 1) sinn-2 x. cos x dx - sinn-1
x cos x ( n 1) ? sinn-2 x.cos2 x dx -
sinn-1 x cos x (n 1) ? sinn-2 x (1 sin2 x)
dx - sinn-1 x cos x (n 1) ? sinn-2 x dx
(n 1) In
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In 1 (n 1) - sinn-1 x cos x ( n 1)
In-2
Therefore In ? sinn x dx

• is the required reduction formula.
• Illustration (i) To find ? sin4 x dx.

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I4 ? sin4 x dx
We need to apply the result (1) again by taking
n 2 That is, I4

I0 ? sin0 x dx ? 1 dx x
Thus I4 ? sin4 x dx

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Illustration (ii) To find ? sin5 x dx
Solution I5 ? sin5 x dx
But I1 ? sin1 x dx - cos x.

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Corollary To evaluate
From (1) , In
But cos ?/2 0 sin 0.
Thus In
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Now, In-2
In
Continuing the process we get
I1 if n is odd.
In
I0 if n is even.
But I1
- cos x0?/2 - (0 1) 1
and I0
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.1 if n is odd.

if n is even.
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Exercise Prove the following
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Evaluation of Integrals
where n is a positive integer.

• We put x sin ?
• Note that when x 0, ? 0 and when x 1, ?
  ?/2.
• we get

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We put x tan ? Note that when x 0, ? 0
and when x ? ?, ? ? ?/2
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Reduction formula for Im,n ? sinm x cosn x dx
Write Im,n ? (sinm-1 x) (sin x cosn x)dx
Then Im,n
- ? (m 1) sinm-2 x cos x
? sinm-2 x cosn x (1 sin2 x) dx

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Evaluation of
Thus we get
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Changing m to m 2 successively, we have

Finally I3,n
if m is odd
I2,n if m
even
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Im,n ? sinm x cosn x dx
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Case (i) When m is odd (and n is even or odd),
Case (ii) When m is even and n is odd,
Case (iii) When m and n are both even,
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Illustrations
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Exercise Prove the following
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Evaluation of Integrals
Put x tan ?, Sol
These values can be computed.
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Put x tan ?, dx sec2 ? d?
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Put x a sin2 ?
Then dx 2a sin ? cos ? d ? ? varies from
0 to ?/2.
a sin ? cos ?.
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Example If n is a positive integer, show that
Solution First we note that
Now we put a x a cos ? .
Then x a (1 cos ?) 2a sin2 (?/2) when x
0, ? 0 and when x 2a, ? ? .
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Reduction formula for In ? tann x dx
In ? (tann-2 x) (tan2 x) dx
? (tann-2 x) (sec2 x 1) dx
? tan n-2 x sec2 x dx - ? tann-2 x dx
This is the reduction formula .
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On changing n to n 2 successively,
The last expression is I1 if n is odd and I0 if
n is even .
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log sec x0?/4
- ..I
where I I1 if n is odd, I I0 if n is even
and I appears with appropriate sign
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