2x-y = 4

y² = 4x and 2x = 4 + y ⇒ 4x = 8 + 2y
y² = 8+2y ⇒ y²-2y-8 = 0
(y-4)(y+2) = 0
y = -2, 4.
Required area \(=\int _{ -2 }^{ 4 }{ ({ x }_{ 1 }-{ x }_{ 2 })dy } \)
\(=\int _{ -2 }^{ 4 }{ \left( \frac { y+4 }{ 2 } -\frac { { y }^{ 2 } }{ 4 } \right) dy } \)
Where x₁ is the line  x₂ is the parabola
\(=\int _{ -2 }^{ 4 }{ \frac { y+4 }{ 2 } dy } -\frac { 1 }{ 4 } \int _{ -2 }^{ 4 }{ { y }^{ 2 }dy } \)
\(=\frac { 1 }{ 2 } { \left[ \frac { { y }^{ 2 } }{ 2 } +4y \right] }_{ -2 }^{ 4 }-\frac { 1 }{ 4 } { \left[ \frac { { y }^{ 3 } }{ 3 } \right] }_{ -2 }^{ 4 }\\ \)
\(=\frac { 1 }{ 2 } \left[ \left( \frac { 16 }{ 2 } +16 \right) -\left( \frac { 4 }{ 2 } -8 \right) \right] -\frac { 1 }{ 4 } \left[ \frac { 64 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ 24+6 \right] -\frac { 1 }{ 4 } \left[ \frac { 64+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } (30)-\frac { 1 }{ 4 } \left( \frac { 72 }{ 3 } \right) \)
\(=15-\frac { 1 }{ 4 } (24)=15-6\)
= 9 sq.units.

y = 3x² - x is an open upward parabola and meets the X-axis at x = 0, x = \(\frac{1}{3}\) when y = 0
∴ The required area is the combination of 3 pies
\(A=\int _{ 1 }^{ 0 }{ ydx+ } \int _{ 0 }^{ \frac { 1 }{ 3 } }{ -ydy } +\int _{ \frac { 1 }{ 3 } }^{ 1 }{ dy } \) [Since the second area lies below the X-axis]
\(A=\int _{ -1 }^{ 0 }{ ({ 3x }^{ 2 }-x)dx+\int _{ 0 }^{ \frac { 1 }{ 3 } }{ \left( x-3{ x }^{ 2 } \right) dx+ } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ ({ 3x }^{ 2 }-x)dx } \)
\(={ \left[ \frac { { 3x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }+{ \left[ \frac { { x }^{ 2 } }{ x } -\frac { { 3x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \frac { 1 }{ 3 } }+{ \left[ \frac { { 3x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } \right] }_{ \frac { 1 }{ 3 } }^{ 1 }\\ \)
\(=0-\left( -1-\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 18 } -\frac { 1 }{ 27 } \right) -0+\left( 1-\frac { 1 }{ 2 } \right) -\left( \frac { 1 }{ 27 } -\frac { 1 }{ 18 } \right) \)
\(=\frac { 3 }{ 2 } +\frac { 1 }{ 18 } -\frac { 1 }{ 27 } +\frac { 1 }{ 2 } -\frac { 1 }{ 27 } +\frac { 1 }{ 18 } \)
\(=\left( \frac { 3 }{ 2 } +\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 18 } +\frac { 1 }{ 18 } \right) -\left( \frac { 1 }{ 27 } +\frac { 1 }{ 27 } \right) \)
\(=2+\frac { 2 }{ 18 } -\frac { 2 }{ 27 } =2+\frac { 1 }{ 9 } -\frac { 2 }{ 27 } =\frac { 54+3-2 }{ 27 } \)
\(A=\frac { 55 }{ 27 } \) sq.units.

Given C'(x) = 2 + 3e3x
\(\Rightarrow \int { C'(x) } =\int { (2+{ 3e }^{ 3x }) } dx\)
\(\Rightarrow C(x)=2x+\frac { { 3e }^{ 3x } }{ 3 } +k\)
\(\Rightarrow C(x)=2x+{ e }^{ 3x }+k\)
Since the fixed cost is Rs. 500, when x = 0, C = 500
\(\Rightarrow 500=0+{ e }^{ 0 }+k\quad [\because { e }^{ 0 }=1]\)
⇒ 500-1 = k ⇒ = 499
∴ C(x) = 2x + e^3x+499
Average cost dunction \(AC=\frac { C }{ x } \)
\(AC=\frac { 2x+{ e }^{ 3x }+499 }{ x } \)
\(=2+\frac { { e }^{ 3x } }{ x } +\frac { 499 }{ x } \)

Given' R'(x) = 7 + e^-0.05x
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (7+{ e }^{ -0.05x })dx } \)
\(={ \left[ 7x+\frac { { e }^{ -0.05x } }{ -0.05 } \right] }_{ 0 }^{ 100 }\)
\(=700-\frac { 100 }{ 5 } ({ e }^{ -5 }-{ e }^{ 0 })\left[ 0.05=\frac { 5 }{ 100 } \right] \)
= 700-20(0.0067-1)[∵e⁰=1]
= 700-0.134
= 719.866
Since the revenue is given in thousands,
Total revenue = 719.866 x 1000
= Rs. 7,19,866

Given \(C'(x)=20+\frac { x }{ 20 } \)
\(\Rightarrow \int { C'(x)dx= } \int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Given when x = 0, C = 200
⇒ k₁ = 200
\(\therefore C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +200\quad ---(1)\)
Also, R'(x) = 30
\(R(x)=\int { 30dx } +{ k }_{ 2 }=30x+{ k }_{ 2 }\)
When x = 0 R = 0 ⇒ K₂ = 0
∴ R(x) = 30x   ----(2)
Profit = Total revenue - total cost
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -200\)
\(\therefore P=10x-\frac { { x }^{ 2 } }{ 40 } -200---(3)\)
\(\frac { dp }{ dx } =10-\frac { 2x }{ 40 } =10-\frac { x }{ 20 } \)
\(\frac { dp }{ dx } =0\)
\(\Rightarrow 10-\frac { x }{ 20 } =0\)
\(\Rightarrow 10=\frac { x }{ 20 } \Rightarrow x=200\)
\(\frac { d^{ 2 }p }{ { dx }^{ 2 } } =-\frac { 1 }{ 20 } <0\)
∴ Profit is maximum when x= 200
∴ maximum profit \(P=10(200)-\frac { { (200) }^{ 2 } }{ 40 } -200[From(3)]\)
P = 2000 -1000 - 200
2000 - 1200
P = Rs .800.
Hence, the maximum profit is Rs. 800.