Given P_d = 18− 2x − x² ; P_s = 2x − 3
We know that at equilibrium prices p_d = p_s
18− 2x − x² = 2x – 3
x² + 4x −21 = 0
(x − 3) (x + 7) = 0
x = –7 or 3
The value of x cannot be negative, x = 3
When x₀ = 3
ஃ p₀ = 18 − 2(3)  − (3)² = 3
CS = \(\int _{ 0 }^{ { x }_{ o } }{ f(x) } \) dx - x₀p₀
= \(\int _{ 0 }^{ 3 }{ (18-2x-{ x }^{ 2 }) } \)dx - 3 x 3
= \({ \left[ 18x-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }\)- 9
= 18(3) - (3)² - \(\left( \frac { { 3 }^{ 3 } }{ 3 } \right) \) - 9
CS = 27 units
PS = x₀P₀ - \(\int _{ 0 }^{ { x }_{ o } }{ g(x) } \)
= (3 \(\times\) 3) - \(\int _{ 0 }^{ 3 }{ (2x-3) } \)
= 9 - \(({ { { x }^{ 2 }-3x) } }_{ 0 }^{ 3 }\)
= 9 units
Hence at equilibrium price,
(i) the consumer’s surplus is 27 units
(ii) the producer’s surplus is 9 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 \(MC=125+10x-\frac { { x }^{ 2 } }{ 9 } \)
\(\Rightarrow \int { MC } =\int { \left( 125+10x-\frac { { x }^{ 2 } }{ 9 } \right) } dx\)
\(\Rightarrow C=125x+\frac { { 10x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 9\times 3 } +k\)
\(\\ \Rightarrow C=125x+{ 5x }^{ 2 }-\frac { { x }^{ 3 } }{ 27 } +k\)
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 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 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