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
\(AC=\frac { C }{ x } =\frac { 100x-{ 5x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } }{ x } +500\)
\(AC=100-5x+\frac { { 0.1x }^{ 2 } }{ 3 } +\frac { 500 }{ x } \)

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

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,
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

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.