C(x) = 10 - 6x
⇒ ഽC(x) = ഽ10.6x dx
\(\Rightarrow C(x)=10.6\frac { { x }^{ 2 } }{ 2 } +{ k }_{ 1 }\)
=5.3x2+k1
Given fixed cost is Rs. 50
When x=s 0, C=50 ⇒ k₁ =50
∴ C(x)  5.3x² + 50 ----(1)
Total revenue =number of units sold x price per unit
∴ Pofit =R(x)-C(x)
=5x-(5.3x²+50)
[From (1)&(2)]
P=5x-5.3x²-50

Given elasticity of demand = \(\frac{p}{x^3}\)
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =\frac { p }{ { x }^{ 3 } } \)
\(\Rightarrow \frac { { -x }^{ 3 }dx }{ x } =p.\frac { dp }{ p } \)
\(\Rightarrow -{ x }^{ 2 }dx=dp\)
\(\Rightarrow -\int { { x }^{ 2 }dx } =\int { dp } \)
\(\Rightarrow -\frac { { x }^{ 3 } }{ 3 } +k=p\)
When p = 2, x = 3
\(\Rightarrow -\frac { { 3 }^{ 3 } }{ 3 } +k=2\)
k = 2 + 9 ⇒ k = 11
∴ (1) becomes
\(P=-\frac { { x }^{ 3 } }{ 3 } +11\)
= 11−\(\frac { { x }^{ 3 } }{ 3 } \)

Given C'(x) = 20 + \(\frac{x}{20}\)
R'(x) = 30
C'(x) = 20 + \(\frac{x}{20}\)
\(\Rightarrow \int { C'(x) } =\int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Since fixed cost is Rs. 100
When x = 0, C = 100 ⇒ k₁ = 100
\(\therefore \ C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +100\) ...(1)
Also, R'(x) = 30
\(\Rightarrow \int { R'(x) } =\int { 30 } dx\)
⇒ R(x) = 30x + k₂
When x = 0, R = 0 ⇒ k₂ = 0
∴ R(x) = 30x ...(2)
Profit function = R(x) - C(x)
\(=30x-\left( 20x+\frac { { x }^{ 2 } }{ 40 } +100 \right) \)
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -100\)
\(P=10x-\frac { { x }^{ 2 } }{ 40 } -100\)

Given MC = 8 + 6x
\(C(x)=\int { (8+6x)dx } \) + k₁
= 8x + 3x² + k_{1 ...}(1)
But given when x = 0, C = 0 ⇒ k₁ = 0
ஃ C(x) = 8x + 3x²  ....(2)
Given that MR = 24
R(x) = \(\int { MR } \) dx + k₂
= \(\int { 24 } \) dx + k₂
= \(\int { 24 } \) + k₂
Revenue = 0, when x = 0 ⇒ k₂ = 0
R(x) = 24x ...(3)
Total Profit functions P(x) = R(x) – C(x)
P(x) = 24x − 8x − 3x²
= 16x − 3x²

y² = 4 ax is the right open
⇒ y = \(\sqrt { 4ax } \) parabola.
The limits are from x = 0 to x = a
∴ Area \(=2\int _{ 0 }^{ a }{ ydx } \)
\(=2\int _{ 0 }^{ a }{ \sqrt { 4ax } } dx\)
\(=2\int _{ 0 }^{ a }{ 2\sqrt { a } \sqrt { x } dx } \)
\(=4\sqrt { a } \int _{ 0 }^{ a }{ \sqrt { x } dx } =4\sqrt { a } \int _{ 0 }^{ a }{ { x }^{ \frac { 1 }{ 2 } }dx } \)
\(=4\sqrt { 3 } { \left[ \frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ a }=4\sqrt { a } \times \frac { 2 }{ 3 } { \left( { x }^{ \frac { 3 }{ 2 } } \right) }_{ 0 }^{ a }\)
\(=8\frac { \sqrt { a } }{ 3 } \left( { a }^{ \frac { 2 }{ 3 } }-0 \right) \)
\(=\frac { 8\sqrt { a } }{ 3 } (a\sqrt { a } )\)
\(=\frac { 8 }{ 3 } { a }^{ 2 }\) sq.units

Area = \(\int _{ 1 }^{ 4 }{ ydx } \)
= \(\int _{ 1 }^{ 4 }{ (4x+3)dx } \)
= \([{ { 2x }^{ 2 }+3x] }_{ 1 }^{ 4 }\) = 32 +12 − 2 − 3
= 39 sq.units

Given MR = 275 - x - 0.3x²
ഽMR = f(275 - x - 0.3x²)dx
To find the total revenue, when it is increased from 10 to 20 units
\(R=\int _{ 10 }^{ 20 }{ (275-x-0.3{ x }^{ 2 })dx } \)
\(={ \left( 275x-\frac { { x }^{ 2 } }{ 2 } -0.3\frac { { x }^{ 3 } }{ 3 } \right) }_{ 10 }^{ 20 }\)
\(={ \left( 275x-\frac { { x }^{ 2 } }{ 2 } -0.1{ x }^{ 3 } \right) }_{ 10 }^{ 20 }\)
\(=\left[ 275(20)-\frac { { 10 }^{ 2 } }{ 2 } -0.1({ 10 }^{ 3 }) \right] \)
= [5500 - 200 - 800] - [2750 - 50 - 100]
= [5500 - 1000] - [2750 - 150]
= 4500-2600 = 1,900
R = Rs. 1,900

Given demand function p = e^-x
and p = 0.5
when p0 = 0.5, 0.5 = e^-x
⇒ \(\frac{1}{2}=e^-x\)
\(\Rightarrow \frac { 1 }{ 2 } =\frac { 1 }{ { e }^{ x } } \Rightarrow ex=2\)
\(\Rightarrow x=\log 2\)
\(\therefore \ { p }_{ 0 }{ x }_{ 0 }=0.5\ \log 2=\frac { 1 }{ 2 } \log 2\)
Consumer's Surplus
\(=\int _{ 0 }^{ x }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ log2 }{ { e }^{ -x }dx-\frac { 1 }{ 2 } log2 } \)
\(={ \left[ \frac { { e }^{ -x } }{ -1 } \right] }_{ 0 }^{ log2 }-\frac { 1 }{ 2 } log2\)
\(=-{ \left[ { e }^{ -x } \right] }_{ 0 }^{ log2 }-\frac { 1 }{ 2 } log2\)
\(=-\left( { e }^{ -log2 }-{ e }^{ 0 } \right) -\frac { 1 }{ 2 } log2\)
\(=-\left( { e }^{ log\quad \frac { 1 }{ 2 } }-1 \right) -\frac { 1 }{ 2 } log2\)
\(=-\left( \frac { 1 }{ 2 } -1 \right) -\frac { 1 }{ 2 } log2\)
\(=-\left( -\frac { 1 }{ 2 } \right) -\frac { 1 }{ 2 } log2\)
\(\\ =\frac { 1 }{ 2 } -\frac { 1 }{ 2 } log2=\frac { 1 }{ 2 } (1-{ log }_{ e }2)\)
∴ C.S = \(\frac{1}{2}\)[1-log_e 2] units

Given y = 36 − x² and y₀ = 11
11 = 36 – x²
x² = 25
x = 5
CS = \(\int _{ 0 }^{ x }{ \text{(demand }\ \text{ function)dx–(Price×quantity demanded)}}\)
= \(\int _{ 0 }^{ 5 }{ (36-{ x }^{ 2 })dx-5\times 11 } \)
= \({ \left[ 36x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 5 }-55\)
= \(\left[ 36(5)-\frac { { 5 }^{ 3 } }{ 3 } \right] -55\)
= \(180-\frac { 125 }{ 3 } -55=\frac { 250 }{ 3 } \)
Hence the consumer’s surplus is = \(\frac { 250 }{ 3 } \)

Given 
\(MR=R'(x)=1500-4x-3{ x }^{ 2 }\)
\(\Rightarrow \int { R'(x) } =\int { (1500-4x-3x^{ 2 })dx } \)
\(\Rightarrow R(x)=1500x-\frac { { 4x }^{ 2 } }{ 2 } -\frac { { 3x }^{ 3 } }{ 3 } +k\)
\(\Rightarrow R(x)=1500x-2{ x }^{ 2 }-{ x }^{ 3 }+k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R(x)=1500x-2{ x }^{ 2 }-{ x }^{ 3 }\)
Average revenue function \(=\frac { R(x) }{ x } \)
\(=\frac { 1500x-{ 2x }^{ 2 }-{ x }^{ 3 } }{ x } \)
AR = 1500 - 2x - x²

Given marginal cost MC = C'(x) = \(\frac { { x }^{ 2 } }{ 200 } +4\)
\(\Rightarrow \int { MC=\int { C'(x)=\int { \left( \frac { { x }^{ 2 } }{ 200 } +4 \right) dx } } } \)
To find the cost of producing 200 air conditioners
\(C=\int _{ 0 }^{ 200 }{ \left( \frac { { x }^{ 2 } }{ 200 } +4 \right) dx } \)
\(={ \left[ \frac { 1 }{ 200 } \times \frac { { x }^{ 3 } }{ 3 } +4x \right] }_{ 0 }^{ 200 }\)
\(={ \left[ \frac { { x }^{ 3 } }{ 600 } +4x \right] }_{ 0 }^{ 200 }\)
\(=\left[ \frac { { (200) }^{ 3 } }{ 600 } +4(200) \right] -0\)
\(=\frac { 8000000 }{ 600 } +800\)
= 13333.33 + 800
= Rs. 14,133.33
Hence, the cost of producing 200 air conditioners is Rs. 14133.33

Given P = Rs.1000, r = \(\frac{5}{100}\) = 0.05 and N = 5
Amount after 5 years
=\(\int _{ 0 }^{ 5 }{ 1000 } { e }^{ 0.05t }dt\)
\(\left[ \therefore A=\int _{ 0 }^{ N }{ p{ e }^{ rl }dt } \right] \)
\(=1000\int _{ 0 }^{ 5 }{ { e }^{ 0.05t }dt } \)
\(=1000{ \left( \frac { { e }^{ 0.05t } }{ 0.05 } \right) }_{ 0 }^{ 5 }\)
\(=\frac { 1000 }{ 0.05 } { \left( \frac { { e }^{ 0.05 } }{ 0.05 } \right) }_{ 0 }^{ 5 }\)
\(=\frac { 1000 }{ 0.05 } { \left( { e }^{ 0.05t } \right) }_{ 0 }^{ 5 }\)
\(=20,000({ e }^{ 0.05(5) }-{ e }^{ 0.05(0) })\)
\(\\ =20,000\left( { e }^{ 0.25 }-{ e }^{ 0 } \right) \)
[∵ e^0.25 = 1.284]
= 20,000(.284-1)
= Rs. 5,680
Hence, the amount after 5 years will be Rs. 5680

Given,
Marginal cost MC = 6 +10x − 6x²
C = \(\int { MC } dx+k\)
= \(\int { (6+10x-{ 6x }^{ 2 })dx+k } \)
= 6x + 5x² − 2x³ + k (1)
when x = 2, C = 12 (given)
12 = 12 + 20 −16 + k
k = -4
C = 6x + 5x² − 2x³ − 4
Average cost = \(\frac { C }{ x } =\frac { 6x+{ 5x }^{ 2 }-2{ x }^{ 3 }+{ 4 } }{ x } \)
= 6 + 5x − 2x² − \(\frac { 4 }{ x } \)

y-1 = x

x	0	-1
y	1	0

Given line is y - 1 = x ⇒ y = x + 1
Given limits are from x = - 2 to 3.
In the diagram, the area from x = - 2 to x = -1 lies below the X-axis and the area from x = -1 to x = 3 lies above the X-axis.
∴ Required Area
\(=\int _{ 2 }^{ -1 }{ -ydx+ } \int _{ -1 }^{ 3 }{ ydx } \)
\(=-\int _{ -2 }^{ -1 }{ ydx } +\int _{ -1 }^{ 3 }{ ydx } \)
\(=\int _{ -1 }^{ -2 }{ ydx } +\int _{ -1 }^{ 3 }{ ydx } \left[ \because \int _{ a }^{ b }{ f(x)dx=-\int _{ b }^{ a }{ f(x)dx } } \right] \)
\(=\int _{ -1 }^{ -2 }{ (x+1)dx+\int _{ -1 }^{ 3 }{ (x+1)dx } } \)
\(={ \left( \frac { { x }^{ 2 } }{ 2 } +x \right) }_{ -1 }^{ -2 }+{ \left( \frac { { x }^{ 2 } }{ 2 } +x \right) }_{ -1 }^{ 3 }\)
\(=\left( \frac { 4 }{ 2 } -2 \right) -\left( \frac { 1 }{ 2 } -1 \right) +\left( \frac { 9 }{ 2 } +3 \right) -\left( \frac { 1 }{ 2 } -1 \right) \)
\(=(2-2)-\left( -1\frac { 1 }{ 2 } \right) +\left( \frac { 15 }{ 2 } \right) -\left( -\frac { 1 }{ 2 } \right) \)
\(=0+\frac { 1 }{ 2 } +\frac { 15 }{ 2 } +\frac { 1 }{ 2 } =\frac { 17 }{ 2 } \) sq.units.

Given MR = 20-5x+3x²
\(\Rightarrow \frac { dR }{ dx } =20-5x+3{ x }^{ 2 }\)
⇒ dR = (20 - 5x + 3x²)dx
⇒ഽdR = ഽ(20-5x+3x²)dx
\(\Rightarrow R=20x-\frac { { 5x }^{ 2 } }{ 2 } +\frac { { 3x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
∴ R = 20x - \(\frac { 5{ x }^{ 2 } }{ x } +{ x }^{ 3 }\)