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 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

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 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 - \(\frac { 5{ x }^{ 2 } }{ x } +{ x }^{ 3 }\)

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.