\(\int { 5dx }\)=5x + c

\(\int { ({ x }^{ 2 } } +x-1)dx=\int { { x }^{ 2 } } dx+\int { xdx } -\int { dx } \)
\(=\frac { { x }^{ 2+1 } }{ 2+1 } +\frac { { x }^{ 1+1 } }{ 1+1 } -x+c\)
\(=\frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } -x+c\)
 

\(\int { 4{x }^{ 3 } } dx=4\int { { x }^{ 3 } } dx\)
\(=4\frac { { x }^{ 3+1 } }{ 3+1 } +c=4\frac { { x }^{ 4 } }{ 4 } +c={ x }^{ 4 }+c\)

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\((i) \ e_{p} =\frac{d x}{d p} \cdot \frac{p}{x}  \)
\(\frac{d x}{d p} =0-4-2 p  \)
\(e_{p} =(-4-2 p) \cdot \frac{2}{\left(30-4 p-p^{2}\right)}  \)
\(=-4-2(2) \cdot \frac{2}{30-4(2)-(2)^{2}}  \)
\(=(-4-4) \times \frac{2}{30-8-4}=\frac{-8 \times 2}{18}=\frac{-16}{18}=-0.89 \)
\((ii) \ T R =P \times Q  \)
\(Q =30-4 p-p^{2}  \)
\(T R =p\left(30-4 p-p^{2}\right)=30 p-4 p^{2}-p^{3}  \)
\(M R =\frac{d R}{d p}=30-8 p-3 p^{2}\)

q = 150 - 3p
TR = P x Q
= P (150 - 3p) = 150 p - 3p²
\(MR={dR\over dx}=150-6p\)

Given the marginal cost function y = 23 + 16x - 3x² ; C = 40
Rs.40 is the fixed cost.
We know that
Total Cost function = \(\int { } \) (Marginal cost function) dx + C
TC=\(\int { } \)ydx+C
=\(\int { } \)(23+16x-3x²)dx+c,where C is a constant,
=\(\int { } \)23dx+\(\int { } \)16x dx - \(\int { } \)3x²dx+C
=23x+16\(\left[ \frac { { x }^{ 2 } }{ 2 } \right] \)-3\(\left[ \frac { { x }^{ 3 } }{ 3 } \right] \)+c
TC=23x+8x²-x³+C
c = 40 given
\(\therefore\) TC = 23x + 8x² - x³ + 40
Average cost function=\(\frac { TC }{ x } \)
=23 + 8x - x² + \(\frac { 40 }{ x } \)

\(TC=20+5{ Q }^{ 2 }+{ 3Q }^{ 3 }\)
\(MC=\frac { d(20) }{ dQ } +\frac { d({ 5Q }^{ 2 }) }{ dQ } +\frac { d(3Q^{ 3 }) }{ dQ } \)
\(=0+5\times 2\quad { Q }^{ 2-1 }+3(3)\quad { Q }^{ 3-1 }\)
\(=0+10Q+9{ Q }^{ 2 }\)
\(MC=10Q+{ 9Q }^{ 2 }\)

To construct the linear supply function at least two points are needed. First data point of supply function is obtained from the statement that the quantity supplied is zero, when the price is Rs 5, that is (0, 5). The second and third data points of supply function can be obtained from the statement that supply increases 10 units for each one rupee rise in price, that is (10, 6) & (20, 7).

The equation of the straight line joining two data points (10, 6) and (20, 7) is given as
The equation of the straight line is
\(\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \)
substituting the values of (X₁, Y₁) (X₂,  Y₂) by (10, 6), (20, 7) respectively,
\(\frac { Y-6 }{ 7-6 } =\frac { X-10 }{ 20-10 }\)
\( \frac { Y-6 }{ 1 } =\frac { X-10 }{ 10 } \)

Then y-6 =  \(\frac { X-10 }{ 10 } \)
10(Y - 6) = X - 10
10Y - 60 = X - 10
10Y - 60 + 10 = X
10Y - 50 = X
-X = -10Y + 50
Multiplying both sides by minus (-), we get
X = -50 + 10Y .
Considering X as quantity supplied and Y as price (P)
Then X = 10p - 50 (or)
(X = -50 + 10p)
If Price = 0; Q = -50
If Q = 0; P = 5
Note: The coefficient of 'p' is - in demand function.
The coefficient of 'p' is + in supply function.

Equation of demand function joining two data points (100, 1) and (50, 2) are (x₁, y₁) and (x₂, y₂) respectively.

\({y-y_1\over y_2-y_1}={x-x_1\over x_2-x_1}\)
\({y-1\over 2-1}={x-100\over 50-100}\)
\({y-1\over 1}={x-100\over -50}\)
Y₁ = 1
Y₂ = 2
X₁ = 100
X₂ = 50
-50 (y - 1)=1 (x -100)
-50y + 50 =x-100
50y = x - 100 - 50
-50y = x-150
-50y + 150 = x
x = 150 - 50y
Hence the demand function is
Q_d = 150 - 50P and Slope M = - 50

TC = 5 + 10² (∴ x = 10)
= 5 + 100 = 105
TC = 105

16x - 4 = 68 + 7x
16x - 7x = 68 + 4
9x = 72;
x = 8

For market equilibrium, P_d = P_s
25 - Q² = 2Q + 1
0 = 25 + Q² + 2Q + 1
0 = -24 + Q² + 2Q
Q² + 2Q - 24 = 0
Q² + 6Q - 4Q - 24 = 0
Q(Q + 6) - 4(Q + 6) = 0
(Q + 6)(Q - 4) = 0
So, Q = 4 or Q =-6. Since Q cannot be equal to -6,
Q = 4
When Q = 4, P_d = 25 - 4² = 9;
P₂ = 2(4)+ 1 = 9
Consumer's surplus =\(\int _{ 0 }^{ 4 }{ (25-Q^{ 2 })dQ)dQ-(9\times 4) } \)
=\({ \left[ 25Q-\frac { Q^{ 3 } }{ 3 } \right] }_{ 0 }^{ 4 }\)-36
=\({ \left[ (25)(4)-\frac { 1 }{ 3 } (4)^{ 3 } \right] }\)-(0)-36
=\(\left[ 100-\frac { 64 }{ 3 } \right] \)-(0)- 36 = 42.67
Producer's surplus Ps
(P_s) = (9 x 4) - \(\int _{ 0 }^{ 4 }{ (2Q+1)dQ } \)
= 36 -\(\left[ (Q^{ 2 }+Q \right] _{ 0 }^{ 4 }\)
= 36 - (16 + 4) = 16

i) zero
\(\mathrm{P} =60-0.2 \mathrm{Q}  \)
\(0 =60-0.2 \mathrm{Q}  \)
\(0.2 \mathrm{Q} =60  \)
\(\mathrm{Q} =\frac{60}{0.2}=\frac{600}{2}=300\)
ii) Rs. 20
\(\mathrm{P} =60-0.2 \mathrm{Q}  \)
\(20 =60-0.2 \mathrm{Q}  \)
\(0.2 \mathrm{Q} =60-20=40 ; \quad \mathrm{Q}=\frac{40}{0.2}=\frac{400}{2}=200\)
iii) Rs. 40
\(\mathrm{P} =60-0.2 \mathrm{Q}  \)
\(40 =60-0.2 \mathrm{Q}  \)
\(0.2 \mathrm{Q} =60-40  \)
\(\mathrm{Q} =\frac{20}{0.2}=\frac{200}{2}=100\)