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7/31/2019 Calculus- Integration
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7/31/2019 Calculus- Integration
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CALCULUS
• Integration is reverse of Differentiation
• Derivative of Function f(x) = f’(x)
•
Integral of Function f’(x) = f(x)
• Integral represents area under Curve
• Integration : defined between Limits is Definite Integration
2
7/31/2019 Calculus- Integration
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CALCULUS• Marginal Cost is given as f’(x) = 3x2 + 2x - 1
• Cost =
= x3 + x2 - x + k
When x= 0 , Constant of Integration = 0
Cost = x3 + x2 - x
Cost of 2 Units = 10
dx x x )1223(
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7/31/2019 Calculus- Integration
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CALCULUS• Marginal Cost Function of a Firm = (logx)2
• Cost Function =
• Integrating by Parts
• C= x (logx)2 -
• C = x (logx)2 - 2{xlogx - x}C = x(logx)2 - { 2xlogx}+ 2x
2)(log x
dx x x
x.log.2
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7/31/2019 Calculus- Integration
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CALCULUS
• If Principal P is paid in the bank every year,
The Interest Rate Compounded every year is i %For a period of n years
A =
If Rs. 48000 is earned uniformly throughout each year
for 5 years with Rate of Interest 12 % . Determine Future
Value of Earnings at end of 5 years
n
indneP
0
..
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CALCULUS
• P = 48000
• i = 0.12
• N = 5 years
• A =
• A =
• A = 48000 { (e 0.12n)/0.12 }
• A = 48000 { (e0.6 / 0.12) - (1/0.12)}
• A = 328844
n
indneP
0
..
ne
12.05
0
.)48000(
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7/31/2019 Calculus- Integration
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CALCULUS• A Bank pays 6 % Interest per annum compounded
continuously. If a person places Rs. 1000 in the bank each
year , how much will be in his account after 3 years.
Ans :
P = 1000
i = 6 %n = 3 years
A =
A =
dneP
n
in..0
dnen ..1000
5
0
06.0
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CALCULUS• A = 1000 ( e 0.06n / 0.06)5
0
A = Rs. 3286.67
Principal explains the money deposited in the bank every year
Number of years indicate the amount of time money is kept
in the bank
Rate of Interest indicates the Discounting Interest in the bank
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CALCULUS• A Sinking Fund is created for redemption of a debenture
of Rs. 100000 at end of 20 years. How much money
should be provided out of profits each year for sinking fundif the Investment Fund can earn interest 9 % per annum.
compounded continuosly.
Ans :
A = 100000 n = 20 years i = 9 %
A =
A =
dneP
n
in..
0
dneP
n
n..
0
09.0
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CALCULUS
•
100000 =
• P = 1782. 50
• P indicates the Principal that should be provided each year
for sinking fund
•
A indicates the Amount
• i indicates the Interest Rate
dxeP
n
x
..0
09.0
10
7/31/2019 Calculus- Integration
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CALCULUS• A machine costs Rs. 55000 and its effective life is
estimated to be 10 years. If Scrap Value is Rs. 5000
only. What should be retained out of profits at end
of each year to accumulate an Interest of 11 % in a manner
that new machine is purchased at same price after 10 years.
Ans : Net Amount of Machine after 10 years
= Total Cost of Machine - Amount Realized From Scrap
= 55000 - 5000
= 50,000
A = n
indneP
0
.
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CALCULUS
• 50,000 =
50,000 =
50,000 = P. {(e0.11x)/ 0.11 )}010
Principal = 2744. 28
dneP
n
in..0
dxePx ..
10
0
11.0
12