Given MC = 2 + 5e^x
\(C=\int { MC } dx+k\)
\(=\int { (2+5{ e }^{ x })dx } +k\)
= 2x + 5e^x + k
Average cost = \(\frac { C }{ x } =\frac { 2x+5{ e }^{ x }+95 }{ x } \)
\(AC=2+\frac { 5{ e }^{ x } }{ x } +\frac { 95 }{ x } \)

Given Curve is y² = 27x³.
\({ x }^{ 3 }=\frac { { y }^{ 2 } }{ 27 } \)
\(x={ \left( \frac { { y }^{ 2 } }{ 27 } \right) }^{ \frac { 1 }{ 3 } }=\frac { { y }^{ \frac { 2 }{ 3 } } }{ 3 } \)
∴ Area \(=\frac { 1 }{ 3 } \int _{ 1 }^{ 2 }{ { y }^{ \frac { 2 }{ 3 } } } dy=\frac { 1 }{ 3 } { \left( \frac { { y }^{ \frac { 2 }{ 3 } +1 } }{ \frac { 2 }{ 3 } +1 } \right) }_{ 1 }^{ 2 }\)
\(={ \left( \frac { 1 }{ 3 } .\frac { { y }^{ \frac { 5 }{ 3 } } }{ \frac { 5 }{ 3 } } \right) }_{ 1 }^{ 2 }\)
\(=\frac { 1 }{ 3 } \times \frac { 3 }{ 5 } { \left( { y }^{ \frac { 5 }{ 3 } } \right) }_{ 1 }^{ 2 }=\frac { 1 }{ 5 } \left( { 2 }^{ \frac { 5 }{ 3 } }-{ 1 }^{ \frac { 5 }{ 3 } } \right) \)
\(=\frac { 1 }{ 5 } \left( { 2 }^{ \frac { 5 }{ 3 } }-1 \right) \)sq.units

Given y = 8x²-4x+6
Area \(=\int _{ 0 }^{ 2 }{ ({ 8x }^{ 2 }-4x+6)dx } \)
\(={ \left( \frac { 8x^{ 3 } }{ 3 } -\frac { { 4x }^{ 2 } }{ 2 } +6x \right) }_{ 0 }^{ 2 }\)
\(={ \left( \frac { 8x^{ 3 } }{ 3 } -{ 2x }^{ 2 }+6x \right) }_{ 0 }^{ 2 }\)
\(=\left[ \frac { 8(8) }{ 3 } -2(4)+6(2) \right] -0\)
\(=\frac { 64+12 }{ 3 } =\frac { 76 }{ 3 } \)
Area = \(\frac { 76 }{ 3 } \)sq.units

Given MR = 6 − 3x² − x³
⇒ ഽMR =ഽ(6 − 3x² − x³)dx
\(\Rightarrow \mathrm{R}=6 x-\frac{\not{3} x^{3}}{\not3}-\frac{x^{4}}{4}+k\)
\(\Rightarrow 6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } \)
Demand function \(P=\frac { R }{ x } =6-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 4 } \)

Given \(MR=\frac { 2 }{ x+3 } -\frac { 2x }{ (x+{ 3) }^{ 2 } } +5\)
\(\Rightarrow \int { MR } =\int { \left( \frac { 2 }{ x+3 } -\frac { 2x }{ (x+{ 3) }^{ 2 } } +5 \right) dx } \)
\(\Rightarrow R=2log(x+3)-2\int { \frac { x }{ { (x+3) }^{ 2 } } dx+5x+k } \)
\(\Rightarrow R=2log(x+3)-2\left[ \int { \frac { x+3-3 }{ { (x+3) }^{ 2 } } dx } \right] +5x+k\)
\(\Rightarrow R=21log(x+3)-2\left[ \int { \frac { 1 }{ x+3 } dx-3\int { \frac { dx }{ { (x+3) }^{ 2 } } } } \right] +5x+k\)

\(\Rightarrow R=\frac { -6 }{ x+3 } +5x+k\)
When x = 0, R = 0 ⇒ k = 0
\(0=\frac { -6 }{ 3 } +0+k\Rightarrow k=+2\)
\(\therefore R=\frac { -6 }{ x+3 } +5x-2\)
Demand function \(P=\frac { R }{ x } \)
\(\Rightarrow P=\frac { -6 }{ x(x+3) } +5+\frac { 2 }{ x } \)
\(\Rightarrow P=\frac { -6+2(x+3) }{ x(x+3) } +5\)
\(\Rightarrow P=\frac { -6+2x+6 }{ x(x+3) } +5\)

\(\Rightarrow P=\frac { 2 }{ x+3 } +5\)

Given \(MC=125+10x-\frac { { x }^{ 2 } }{ 9 } \)
\(\Rightarrow \int { MC } =\int { \left( 125+10x-\frac { { x }^{ 2 } }{ 9 } \right) } dx\)
\(\Rightarrow C=125x+\frac { { 10x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 9\times 3 } +k\)
\(\\ \Rightarrow C=125x+{ 5x }^{ 2 }-\frac { { x }^{ 3 } }{ 27 } +k\)
Given fixed cost is Rs. 250,
When x = 0, C = 250 ⇒ k = 250
\(\therefore C=125x+{ 5x }^{ 2 }-\frac { { x }^{ 3 } }{ 27 } +250\)
When x = 15, C = ?
\(\therefore \ C=125(15)+5{ (15) }^{ 2 }-\frac { { 15 }^{ 3 } }{ 27 } +250\)
\(=1875+1125-\frac { 3375 }{ 27 } +250\)
= 3000-125+250
= Rs. 3125
∴ C = Rs. 3125

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_d = 1600 - x² and
Supply function P_s = 2x² + 400
Under perfect competition p_d = p_s
⇒ 1600 - x² = 2x² + 400
⇒ 1600-400 = 2x² + x²
⇒ 1200 = 3x²
\(\Rightarrow \frac { 1200 }{ 3 } ={ x }^{ 2 } \Rightarrow { x }^{ 2 }=400\)
\(\Rightarrow x=\pm \sqrt { 400 } =+20\quad or-20\)
Since x cannot be negative, x₀ = 20
∴ p₀ = 1600 - (20)² = 1600 - 400
= 1200
∴ p₀x₀ = (1200) (20) = 24000
Producer's Surplus PS = P_ox_o -\(\int _{ 0 }^{ x }{ g(x)dx } \)
\(=24000-\int _{ 0 }^{ 20 }{ ({ 2x }^{ 2 }+400)dx } \)
\(=24000-{ \left[ \frac { { 2x }^{ 3 } }{ 3 } +400x \right] }_{ 0 }^{ 20 }\)
\(=24000-\left[ \frac { { 2(20) }^{ 3 } }{ 3 } +400(20) \right] \)
\(=24000-\left[ \frac { 16000 }{ 3 } +8000 \right] \)
\(=24000-\left[ \frac { 16000+24000 }{ 3 } \right] \)
\(=24000-\left[ \frac { 40000 }{ 3 } \right] \)
\(=\frac { 72000-40000 }{ 3 } \)
PS = \( \frac{32000}{3}\)units

Given demand function p = \(\frac{36}{x+4}\)
Also, given market price is 6
⇒ p₀ = 6
\(\therefore \ \frac{1}{\not 6}=\frac {\not 36}{x_{0}+4}\)
⇒ x₀ + 4 = 6
⇒ x0 = 6 - 4 = 2
⇒ x₀ = 2
⇒ p₀x₀ = 6 x 2 = 12
Consumer's Surplus
\(CS=\int _{ 0 }^{ x }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \frac { 36 }{ x+4 } dx-12 } \)
\(=36 \log{ \left[ x+4 \right] }_{ 0 }^{ 2 }-12\)
= 36 [log(2 + 4) -log(0 +4)] - 12
= 36 [log 6 -log 4] - 12
= 36log\((\frac{6}{4})\)-12
[∴ log m-log n = log \((\frac{m}{n})\)]
CS = 36 log\((\frac{3}{2})\)-12 units

Given supply function P = 3x + 5x² and x = 4.
When x₀ = 4, p₀ = 3(4)+5(4²)
= 12 + 80 = 92
∴ p₀x₀ = 92 x 4 = 368
Producer's Surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x }{ g(x)dx } \)
\(=368-\int _{ 0 }^{ 4 }{ (3x+5{ x }^{ 2 })dx } \)
\(=368-{ \left( \frac { { 3x }^{ 2 } }{ 2 } +\frac { { 5x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 4 }\)
\(=368-\left( \frac { 3({ 4 }^{ 2 }) }{ 2 } +5\frac { ({ 4 }^{ 3 }) }{ 3 } \right) \)
\(=368-\left( 24+\frac { 320 }{ 3 } \right) \)
= 368-(24+106.66)
= 368-(130.66)
P.S = 237.3 units

Given supply function p = 7 + x and x = 5
When x₀ = 5, p₀ = 7 + 5 = 12
∴ p₀x₀ = x₁₂ = 60
Producer's Surplus
\(={ p }_{ 0 }{ x }_{ 0 }-\int _{ g }^{ x }{ (x)dx } \)
\(=60-\int _{ 0 }^{ 5 }{ (7+x)dx } \)
\(=60-{ \left[ 7x+\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 5 }\)
\(=60-\left[ 7(5)+\frac { { 5 }^{ 2 } }{ 2 } \right] \)
\(=60-\left[ 35+\frac { 25 }{ 2 } \right] \)
= 60-(35+12.5)
= 60-47.5
P.S = 12.5 = \(\frac{25}{2}\) units.

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 demand function P_d = 85 - 5x and 
Supply function p₅ = 3x - 35
At equilibrium prices,P_d = P_s
⇒ 85 - 5x = 3x - 35
⇒ 85 + 35 = 3x + 5x
⇒ 120 = 8x
⇒ x = \(\frac{120}{8}\) = 15
When x₀ = 15, p₀ = 85-5(15)
= 85 - 75 = 10
p₀ = 10
∴ p₀x₀ = 15\(\times\)10 = 150
Consumer's Surplus
\(Cs=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 15 }{ (85-5x)dx-150 } \)
\(={ \left[ 85x-\frac { 5{ x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 15 }-150\)
\(=85(15)-5\frac { { (15) }^{ 2 } }{ 2 } -150\)
\(=1275-\frac { 1125 }{ 2 } -150\)
= 1275 - 562.5 - 150
= 1275 - 712.5
Cs = 562.5

Given demand functionp = 122 - 5x - 2x² and x = 6
When x₀ = 6, p₀ = 122-5(6)-2(6)²
= 122-30-72
= 122-102
P₀ = 20
p₀x₀ = 20 \(\times\) 6 = 120
Consumer's Surplus
CS \(=\int _{ 0 }^{ x }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(=\int _{ 0 }^{ 6 }{ (122-5x-2{ x }^{ 2 })dx-120 } \)
\(={ \left[ 122x-\frac { 5{ x }^{ 2 } }{ 2 } -\frac { { 2x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 6 }-120\)
\(=122(6)-5\frac { \left( { 6 }^{ 2 } \right) }{ 2 } -2\frac { \left( { 6 }^{ 3 } \right) }{ 3 } -120\)
\(=732-\frac { 180 }{ 2 } -\frac { 432 }{ 3 } -120\)
= 732-90-144-120
= 732-354
C.S = 378 units

Given demand function P = 50 - 2x
and x = 20
when x₀ = 20,
p₀ = 50 - 2(20) = 50-40 = 10
p₀x₀ = 20(10) = 200
Consumer's Surplus \(CS=\int _{ 0 }^{ x0 }{ f(x) } dx-{ p }_{ 0 }{ x }_{ 0 }\)
\(CS=\int _{ 0 }^{ 20 }{ (50-2x)dx-200 } \)
= 50(20)-(20)²-200
= 1000-400-200
= 1000-600
cs = 400 units

g(x) = 4x + 8 and x₀ = 5
p₀ = 4(5) + 8 = 28
PS = x₀ p₀ – \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \) dx
= (5 × 28) - \(\int _{ 0 }^{ 5 }{ (4x+8) } \) dx
= 140 – \({ \left[ 4\left( \frac { { x }^{ 2 } }{ 2 } \right) +8x \right] }_{ 0 }^{ 5 }\)
= 140 – (50 + 40)
= 50 units
Hence the producer’s surplus = 50 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 = 14 - x + 9x²
⇒ \(\frac{dR}{dx}\) = 14 - 6x + 9x²
⇒ dR = (14 - 6x + 9x²)dx
⇒ ഽdR = ഽ(14 - 6x + 9x²)dx
\(\Rightarrow \ R=14x-\frac { 6{ x }^{ 2 } }{ 2 } +\frac { { 9x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
R = 14x - 3x²+ 3x³
Demand function \(P=\frac { R }{ x } =\frac { 14x-3{ x }^{ 2 }+{ 3x }^{ 3 } }{ x } \)
⇒ P = 14 − 3x + 3x²

Given \(MC=\frac { 14000 }{ \sqrt { 7x+4 } } \)
\(\Rightarrow \frac { dc }{ dx } =\frac { 14000 }{ \sqrt { 7x+4 } } \)
\(\Rightarrow dC=\frac { 14000 }{ \sqrt { 7x+4 } } dx\)
\(\Rightarrow \int { dC } =14000\int { \frac { dx }{ \sqrt { 7x+4 } } } \)
\(\Rightarrow 14000\int { (7x+4)^{ -\frac { 1 }{ 2 } }dx } \)
\(\Rightarrow C=14000\frac { { (7x+4) }^{ -\frac { 1 }{ 2 } +1 } }{ \left( -\frac { 1 }{ 2 } +1 \right) 7 } +k\)
\(\Rightarrow C=14000\frac { { (7x+4) }^{ \frac { 1 }{ 2 } } }{ \frac { 7 }{ 2 } } +k\)
\(C={\not14000} \times \frac{2}{\not 7} \sqrt{7 x+4}+k\)
\(\Rightarrow C=4000\sqrt { 7x+4 } +k\)  ...(1)
Given fixed cost is 18,000
When x = 0, C = 18000
⇒ 18000 = 4000√4 + k
⇒ 18000 = 4000(2) + k
⇒ k = 18000-8000
⇒ k = 10000
∴(1) becomes,
C = 4000\(\sqrt { 7x+4 } +10000\), 
Average Cost (AC) \(=\frac{C}{x}\)
= \(\frac { 4000 }{ x } \sqrt { 7x+4 } +\frac { 10000 }{ x } \)

Given
\(MR=10+3x-{ x }^{ 2 }\)
\(\frac { dR }{ dx } =10+3x-{ x }^{ 2 }\)
\(\Rightarrow dR=(10+3x-{ x }^{ 2 })dx\)
\(\Rightarrow \int { dR } =\int { (10+3x-{ x }^{ 2 }) } dx\)
\(\Rightarrow R=10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 2 } }{ 3 } \)
Demand function\(P=\frac { R }{ x } =\frac { 10x+\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } }{ x } \)
P = 10 + \(\frac { { 3x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \)

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 revenue function \(=\frac { 4 }{ { (2x+3) }^{ 2 } } -1\)
\(\Rightarrow MR=\frac { dR }{ dx } =4(2x+3{ ) }^{ -2 }-1\)
\(\Rightarrow dR=(4(2x+3{ ) }^{ -2 }-1)dx\)
\(\Rightarrow \int { dR=4\int { (2x+3{ ) }^{ -2 } } dx-\int { dx } } \)
\(\Rightarrow R=4\frac { { (2x+3) }^{ -2+1 } }{ { (-2+1) }^{ 2 } } -x\)
\(\Rightarrow R=4\frac { (2x+3{ ) }^{ -1 } }{ -2 } -x\)
\(\Rightarrow R=\frac { -2 }{ (2x+3) } -x+k\) ..(1)
When x = 0, R = 0
\(\Rightarrow 0=\frac { -2 }{ 3 } -0+k\Rightarrow =\frac { 2 }{ 3 } \)
∴ (1) becomes 
\(R=\frac { -2 }{ 2x+3 } -x+\frac { 2 }{ 3 } \)
We know that R = Px ⇒ P = \(\frac{R}{x}\)
∴ Average revenue function
\(P=\frac { R }{ x } =\frac { R }{ x } =\frac { -2 }{ x(2x+3) } -\frac { x }{ x } +\frac { 2 }{ 3x } \)
\(=\frac { -2 }{ x(2x+3) } -1+\frac { 2 }{ 3x } \)
\(=\frac { -2 }{ x(2x+3) } +\frac { 2 }{ 3x } -1\)
\(=\frac { -6+2(2x+3) }{ 3x(2x+3) } -1\)
\(=\frac{-\not 6+4 x+\not 6}{3 x(2 x+3)}-1 \)
\(=\frac{4 \not x}{3\not x(2 x+3)}-1\)
\(=\frac { 4 }{ 3(2x+3) } -1\)
\(\\ \\ P=\frac { 4 }{ 6x+9 } -1\)
Hence Proved.

Given marginal revenue
MR = 9-4x²
\(\Rightarrow \frac { dR }{ dx } =9-4{ x }^{ 2 }\) 
\(\Rightarrow \int { dR } =\int { (9-4{ x }^{ 2 })dx } \)
\(\Rightarrow R=9x-\frac { { 4x }^{ 3 } }{ 3 } +k\quad ...(1)\)
When no product is sold, revenue is zero
∴ When x = 0, R = 0
⇒0 = 0 - 0 + k ⇒ k = 0
∴ (1) becomes 
\(R=9x-\frac { 4{ x }^{ 3 } }{ 3 } \)
We know that R = Px where P is the demand function
\(\Rightarrow \mathrm{P} \not x=\not x\left(9-\frac{4 x^{2}}{3}\right)\)
\(\Rightarrow P=9-\frac { 4{ x }^{ 3 } }{ 3 } \)
Hence, the demand function is \(9-\frac { 4{ x }^{ 3 } }{ 3 } \)

Given marginal revenue MR = 5 + 3e - 0.03x
\(\Rightarrow \frac { dR }{ dx } =5+{ 3e }^{ -0.03x }\)
\(\Rightarrow dR=(5+3{ e }^{ -0.03x })dx\)
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (5+3{ e }^{ -0.03x })dx } \)
\(={ \left[ 5x+3\frac { { e }^{ -0.03x } }{ -0.03 } \right] }_{ 0 }^{ 100 }\)
\(={ \left[ 5x-\frac { { e }^{ -0.03x } }{ 0.01 } \right] }_{ 0 }^{ 100 }\)
\(=\left( 5(100)-\frac { { e }^{ -0.03(100) } }{ 0.01 } \right) -\left( 0-\frac { { e }^{ -0.03(0) } }{ 0.01 } \right) \)
\(=500-\frac { { e }^{ -3 } }{ .01 } +\frac { { e }^{ 0 } }{ 0.01 } \)
\(=500-\frac { .05 }{ .01 } +\frac { 1 }{ .01 } \)
= 500 - 5 + 100
= 600 - 5 = 595
∴ Total revenue = 595\(\times\)1000
= Rs. 595000
[∵ Revenue is given in thousands]

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 marginal cost function = \(\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow MC=\frac { a }{ \sqrt { ax+b } } \Rightarrow \frac { dC }{ dx } =\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow dC=\frac { a }{ \sqrt { ax+b } } dx\)
\(\Rightarrow \int { dC } =a\int { \frac { dx }{ \sqrt { ax+b } } } \)
\(\Rightarrow C=a\int { { (ax+b) }^{ \frac { -1 }{ 2 } }dx } \)
\(\Rightarrow C=\not a \frac{(a x+b)^{\frac{-1}{2}+1}}{\left(-\frac{1}{2}+1\right) \not a}+k\)
\(\Rightarrow C=\frac { { (ax+b) }^{ \frac { 1 }{ 2 } } }{ 1 } +k\)
\(\Rightarrow C=2\sqrt { ax+b } +k\) ...(1)
Since the cost of output is zero.
C = 0, when x = 0
\(\therefore (1)\rightarrow 0=2\sqrt { 0+b } +k\)
\(\Rightarrow 0=2\sqrt { b } +k\)
\(\Rightarrow k=-2\sqrt { b } \)
∴(1)becomes
\(C=2\sqrt { ax+b } -2\sqrt { b } \)

Given \(MC=300{ x }^{ \frac { 2 }{ 5 } }\)
\(\Rightarrow \frac { dC }{ dx } =300{ x }^{ \frac { 2 }{ 5 } }\)
\(\int { dC } =300\int { { x }^{ \frac { 2 }{ 5 } }dx } \)
\(\Rightarrow C=300\frac { { x }^{ \frac { 2 }{ 5 } +1 } }{ \frac { 2 }{ 5 } +1 } +k\)
\(\Rightarrow C=300\frac { { x }^{ \frac { 7 }{ 5 } } }{ \frac { 7 }{ 5 } } +k\)
\(\Rightarrow C=300\times \frac { 5 }{ 7 } { x }^{ \frac { 7 }{ 5 } }+k\) ...(1)
Given fixed cost is zero ⇒ k = 0
∴(1) becomes,
\(C=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } }+0\Rightarrow C=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } }\)
Average cost function \((AC)=\frac { C }{ x } \)
\(\Rightarrow AC=\frac { 1500 }{ 7 } \frac { { x }^{ \frac { 7 }{ 5 } } }{ x } \)
\(\Rightarrow AC=\frac { 1500 }{ 7 } { x }^{ \frac { 7 }{ 5 } -1 }\)
\(\Rightarrow AC=\frac { 1500 }{ 7 } { x }^{ \frac { 2 }{ 5 } }\)

Given marginal cost function
\(MC=\frac { dc }{ dx } =100-10x+0.1{ x }^{ 2 }\)
\(\Rightarrow C=\int { (100-10x+0.1{ x }^{ 2 })dx } \)
\(\Rightarrow C=100x-\frac { { 10x }^{ 2 } }{ 2 } +\frac { { 0.1x }^{ 3 } }{ 3 } +k\) ...(1)
Given fixed cost is Rs. 500 ⇒ k = 500
∴ (1) becomes,
\(C=100x-5{ x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } +500\)
Average cost function
\(AC=\frac { C }{ x } =\frac { 100x-{ 5x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } }{ x } +500\)
\(AC=100-5x+\frac { { 0.1x }^{ 2 } }{ 3 } +\frac { 500 }{ x } \)

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 inventory on hand I(x) = 500 - 0.03x²
Holding cost per scooter is C₁ = Rs. 0.3
Time period T = 30 days
Total inventory carrying cost
\(={ C }_{ 1 }\int _{ 0 }^{ r }{ I(x)dx } \)
\(=0.3\int _{ 0 }^{ 30 }{ ({ 500-0.03x }^{ 2 })dx } \)
\(=0.3{ \left[ 500x-\frac { { 0.03x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 30 }\)
\(=0.3{ \left[ 500x-0.01{ x }^{ 3 } \right] }_{ 0 }^{ 30 }\)
\(=0.3\left\{ \left[ 500(30)-0.01{ (30) }^{ 3 } \right] -0 \right\} \)
= 0.3[15000 - 0.01 (27000)]
= 0.3[15000 - 270]
= 0.3 (14730)
= Rs. 4419
Hence, the total cost or maintaining inventory for 30 days = Rs. 4419.

Annual consumption at timet = 0 (In the year 2000) = p₀ = 2547 metric ton.
Total production of Gold from 2000 to 2013 = \(\int _{ 0 }^{ 13 }{ 2547e^{ 0.006t } } dt\)
= \(\frac { 2547 }{ 0.006 } \left[ e^{ 0.006t } \right] _{ 0 }^{ 13 }\)
= 424500 (e^0.078 −1)
= 34,426.95 metric tons approximately.

Given cost of overhaul of an engine is Rs. 10,000
Operating cost per hour = 2x - 240.
Total cost for the engine to run for 300 hours
after overhaul =10,000+\(\int _{ 0 }^{ 300 }{ (2x-240) } dx\)
\(=10,000+{ \left[ \frac { { 2x }^{ 2 } }{ 2 } -240x \right] }_{ 0 }^{ 300 }\)
\(=10,000+{ [{ x }^{ 2 }-240x] }_{ 0 }^{ 300 }\)
= 10,000 + 90,000 - 72,000
= 1,00,000 - 72,000
= Rs. 28,000

p = 10000, r = 0.1, N = 5
Annuity = \(\int _{ 0 }^{ 5 }{ 10000 } { e }^{ 0.1t }\ dt\)
= \(\frac { 10000 }{ 0.1 } { \left( { e }^{ 0.1t } \right) }_{ 0 }^{ 5 }\)
= \(100000[{ e }^{ 0.1\times 5 }-{ e }^{ 0 }]\)
= 100000 (e^0.5 −1)
=100000 [0.6487]
= Rs. 64,870

Here I(x) = 200 – 0.2x
C₁ = Rs. 3.5
T = 30
Total inventory carrying cost = C₁\(\int _{ 0 }^{ r }{ I(x) } dx=3.5\int _{ 0 }^{ 30 }{ (200-0.2x) } dx\)
= \(3.5{ \left( 200x-\frac { { 0.2x }^{ 2 } }{ 2 } \right) }_{ 0 }^{ 30 }\) = 20,685

Given MR = 35 + 7x − 3x²
R = \(\int { (MR)dx } +k\)
= \(\int { (35+7x+3{ x }^{ 2 }) } dx+k\)
R = 35x + \(\frac { 7 }{ 2 } \) x² − x³ + k
Since R = 0 when x = 0, k = 0
R = 35x + \(\frac { 7 }{ 2 } \) x² − x³ 
Demand function P = \(\frac{R}{x}\)
P = 35 + \(\frac { 7 }{ 2 } \) x - x²

Saving Cost S(t) = \(\int _{ 0 }^{ t }{ 20000t } \ dt\)
= 10000 t²
To recoup the total price,
10000 t² = 640000
t² = 64
t = 8
When t = 8 years, one can recoup the price.

Let p be the additional product produced for additional of x labour,
\(\frac{dp}{dx}\) = 300 - 5x^2/3
\(p=\int _{ 0 }^{ 64 }{ { \left( 300-5{ x }^{ \frac { 2 }{ 3 } } \right) } } dx\)
\(={ \left[ 300x-3{ x }^{ \frac { 5 }{ 3 } } \right] }_{ 0 }^{ 64 }\)
= 300 \(\times\) 64 -3(64)^5/3
= 16128
∴ The number of additional units produced 16128
Total number of units produced by 264 workers
= 50,000 + 16,128 = 66128 units

Total sale = \(\int _{ 0 }^{ 4 }{ (100-{ 90 }e^{ -x }) } dx\)
= \({ (100x+90e^{ -x }) }_{ 0 }^{ 4 }\)
= 400 + 90e⁻⁴ −(0 + 90)
= 400 + 90(0.018) −90
= 311.62 units

Given MC = 2 + 5e^x
\(C=\int { MC } dx+k\)
\(=\int { (2+5{ e }^{ x })dx } +k\)
= 2x + 5e^x + k
x = 0 ⇒ C = 100,
100 = 2(0)+ 5(e⁰ )+ k
k = 95
C = 2x + 5 e^x + 95.

\(MC=125+10x-\frac { { x }^{ 2 } }{ 9 } \)
\(C=\int { MC } dx+k\)
\(=\int { \left( 125+10x-\frac { x^{ 2 } }{ 9 } \right) } dx+k\)
= 125 + 5x²- \(\frac { { x }^{ 3 } }{ 27 } \) + k
Fixed cost k = 250
C = 125x + 5x²- \(\frac { { x }^{ 3 } }{ 27 } \) + 250
When x = 15
C = 125(15) + 5(15)²- \(\frac { { (15) }^{ 3 } }{ 27 } \) + 250
= 1875 +1125 −125 + 250
C = Rs. 3,125

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 − \(\frac { 4 }{ x } \)

Given curve y = 4x² is an open upward parabola
\(\Rightarrow \frac { y }{ 4 } ={ x }^{ 2 }\)
The limits are from y = 0 to y = 4
Since the shaded region lies to the right of Y-axis,
required area \(=\int _{ 0 }^{ 4 }{ xdy } \)
\(=\int _{ 0 }^{ 4 }{ \sqrt { \frac { y }{ 4 } } dy } \)
\(=\frac { 1 }{ 2 } \int _{ 0 }^{ 4 }{ \sqrt { y } dy } =\frac { 1 }{ 2 } \int _{ 0 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)
\(=\frac{1}{2}\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4}=\frac{1}{\not2} \times \frac{\not2}{3}\left[y^{\frac{3}{2}}\right]_{0}^{4}\)
\(=\frac { 1 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } }-0 \right] =\frac { 1 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 1 }{ 3 } { 4 }^{ 1 }\sqrt { 4 } =\frac { 1 }{ 3 } .4(2)\)
A = \(\frac{8}{3}\) sq.units

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.

y = x

x	1	2
y	1	2

Given line y = x
The limits are x = 1, x = 2
Since the shaded area lies to the right of Y-axis,
Area \(=\int _{ 1 }^{ 2 }{ y\quad dx } \)
\(=\int _{ 1 }^{ 2 }{ x\quad dx } ={ \left( \frac { { x }^{ 2 } }{ 2 } \right) }_{ 1 }^{ 2 }\)
\(=\frac { { 2 }^{ 2 } }{ 2 } -\frac { { 1 }^{ 2 } }{ 2 } =2-\frac { 1 }{ 2 } \)
= \(\frac{4-1}{2}=\) \(\frac{3}{2}\) sq.units

Required area = \(\int _{ -1 }^{ 0 }{ -xdx } +\int _{ 0 }^{ 2 }{ xdx } \)
\(=-{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }{ +\left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 2 }=-\left[ 0-\frac { 1 }{ 2 } \right] +\left[ \frac { 4 }{ 2 } -0 \right] \)
\(=\frac { 5 }{ 2 } \) sq.units