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.

The equation y² = 16x represents a parabola (Open rightward)
Required Area = 2\(\int _{ a }^{ b }{ y } dx\)
\(=2\int _{ 0 }^{ 4 }{ \sqrt { 16x } } \ dx\)
\(=8\int _{ 0 }^{ 4 }{ { x }^{ \frac { 1 }{ 2 } } } dx=8{ \left[ { \frac { { x }^{ { \frac { 3 }{ 2 } } } }{ \frac { 3 }{ 2 } } } \right] }_{ 0 }^{ 4 }=\frac { 16 }{ 3 } \left( { \left( 4 \right) }^{ \frac { 3 }{ 2 } } \right) =\frac { 128 }{ 3 } \) sq.units

Equation of the required circle is \({ x }^{ 2 }+{ y }^{ 2 }={ a }^{ 2 }\) (1)
put \(y=0\), \({ x }^{ 2 }={ a }^{ 2 }\)
⇒  \(x=\pm a\)
Since equation (1) is symmetrical about both the axes
The required area = 4 [Area in the first quadrant between the limit 0 and a.]
\(=4\int _{ 0 }^{ a }{ y } \ dx\)
\(=4\int _{ 0 }^{ a }\sqrt { { a }^{ 2 }-{ x }^{ 2 } } dx=4{ \left[ \frac { x }{ 2 } \sqrt { { a }^{ 2 }-{ x }^{ 2 } } +\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{ \frac { x }{ a } } \right] }_{ 0 }^{ a }\)
\(=4{ \left[ 0+\frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{( \frac { a }{ a } )} \right] }=4{ \left[ \frac { { a }^{ 2 } }{ 2 } \sin ^{ -1 }{( 1)} \right] }=4.\frac { { a }^{ 2 } }{ 2 }\frac { {π} }{ 2 }\)
= πa² sq. units

\(y=\left| x+3 \right| =\begin{cases} x+3\quad if\quad x\ge -3\quad \\ -(x+3)\quad if\quad x<-3 \end{cases}\)
Required area = \(\int _{ b }^{ a }{ y } dx=\int _{ -6 }^{ 0 }{ y } dx\)
= \(\int _{ -6 }^{ -3 }{y}\ dx+\int _{ -3 }^{ 0 }{ y} dx\)
= \(\int _{ -6 }^{ -3 }{ -(x+3) } dx+\int _{ -3 }^{ 0 }{ (x+3) } dx\)
= \(-{ \left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -6 }^{ -3 }{ +\left[ \frac { { (x+3) }^{ 2 } }{ 2 } \right] }_{ -3 }^{ 0 }\) 
\(=-\left[ 0-\frac { 9 }{ 2 } \right] +\left[ \frac { 9 }{ 2 } -0 \right] \)
= 9 sq. units

 

Area = \(\int _{ 1 }^{ 4 }{ ydx } \)
= \(\int _{ 1 }^{ 4 }{ (4x+3)dx } \)
= \([{ { 2x }^{ 2 }+3x] }_{ 1 }^{ 4 }\) = 32 +12 − 2 − 3
= 39 sq.units

\(y=4{ - }x^{ 2 }\)
Required area = \(\int _{ 0 }^{ 2 }{ ydx } =\int _{ 0 }^{ 2 }{ (4- } { x }^{ 2 })dx\)
= \({ \left[ 4x-\frac { { x }^{ 2 } }{ 3 } \right] }_{ 0 }^{ 2 }=8-\frac { 8 }{ 3 } \)
= \(\frac { 16 }{ 3 } \) sq.units

x - 2y - 12 = 0
x = 2y + 12
Required Area
= \(\int _{ 2 }^{ 5 }{ xdy } \)
= \(\int _{ 2 }^{ 5 }{ (2y+12)dy= } [{ { y }^{ 2 }+12y] }_{ 2 }^{ 5 }\)
= (25 + 60)−(4 + 24) = 57 sq.units

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 = \(\frac { 250 }{ 3 } \)

Required area = \(\int _{ -1 }^{ 0 }{ -xdx } +\int _{ 0 }^{ 2 }{ xdx } \)
\(=-{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }{ +\left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 2 }=-\left[ 0-\frac { 1 }{ 2 } \right] +\left[ \frac { 4 }{ 2 } -0 \right] \)
\(=\frac { 5 }{ 2 } \) sq.units