y = x² + 2
y-2 = x²
It is a parabola open upward with vertex (0, 2)
\( \text {Area } =\int_{a}^{b} y d x \)
\( =\int_{0}^{3}\left(y_{1}-y_{2}\right) \)
y₁ \(\Rightarrow\) Area underthe parabola y = x²+2
y₂ \(\Rightarrow\) Area under the line y = x
\( =\int_{0}^{3}\left(x^{2}+2-x\right) d x=\left[\frac{x^{3}}{3}+2 x-\frac{x^{2}}{2}\right]_{0}^{3} \)
\( =9+6-\frac{9}{2}=15-\frac{9}{2} \)
\( =\frac{21}{2} \text { sq. units }\)

= {-(x-5) if x<0
= {+(x-5) if x>0
Required area \(=\int _{ 0 }^{ 1 }{ |x-5|dx } \)
\(=\int _{ 0 }^{ 1 }{ -(x-5)dx } \)
\(=\int _{ 0 }^{ 1 }{ (5-x)dx } \)
\(={ \left[ 5x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=\left[ 5(1)-\frac { 1 }{ 2 } \right] -0\)
\(=5-\frac { 1 }{ 2 } =\frac { 10-1 }{ 2 } \)
\(=\frac { 9 }{ 2 } \) sq.units.

Required area \(=\int _{ 0 }^{ \pi }{ ydx } \)
\(=\int _{ 0 }^{ \pi }{ ydx } +\int _{ \pi }^{ 2\pi }{ -ydx } \)
[∴ the second area lies below the X-asis]
\(=\int _{ 0 }^{ \pi }{ sin\quad xdx } +\int _{ \pi }^{ 2\pi }{ -sin\quad xdx } \)
\(={ \left[ -cos\quad x \right] }_{ 0 }^{ \pi }-{ \left[ -cos\quad x \right] }_{ \pi }^{ 2\pi }\)
\(=-{ \left[ cos\quad x \right] }_{ 0 }^{ \pi }+{ \left[ cos\quad x \right] }_{ \pi }^{ 2\pi }\)
\(=-\left[ cos\pi -cos0 \right] +\left[ cos2\pi -cos\pi \right] \)
=-[-1-1]+[+1-(-1)]
\([\therefore cos\pi =-1cos\quad 0=1\& \quad cos\quad 2\pi =1]\)
=-(-2)+(2)
=2+2
A=4sq.units.

2x-y = 4

y² = 4x and 2x = 4 + y ⇒ 4x = 8 + 2y
y² = 8+2y ⇒ y²-2y-8 = 0
(y-4)(y+2) = 0
y = -2, 4.
Required area \(=\int _{ -2 }^{ 4 }{ ({ x }_{ 1 }-{ x }_{ 2 })dy } \)
\(=\int _{ -2 }^{ 4 }{ \left( \frac { y+4 }{ 2 } -\frac { { y }^{ 2 } }{ 4 } \right) dy } \)
Where x₁ is the line  x₂ is the parabola
\(=\int _{ -2 }^{ 4 }{ \frac { y+4 }{ 2 } dy } -\frac { 1 }{ 4 } \int _{ -2 }^{ 4 }{ { y }^{ 2 }dy } \)
\(=\frac { 1 }{ 2 } { \left[ \frac { { y }^{ 2 } }{ 2 } +4y \right] }_{ -2 }^{ 4 }-\frac { 1 }{ 4 } { \left[ \frac { { y }^{ 3 } }{ 3 } \right] }_{ -2 }^{ 4 }\\ \)
\(=\frac { 1 }{ 2 } \left[ \left( \frac { 16 }{ 2 } +16 \right) -\left( \frac { 4 }{ 2 } -8 \right) \right] -\frac { 1 }{ 4 } \left[ \frac { 64 }{ 3 } +\frac { 8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } \left[ 24+6 \right] -\frac { 1 }{ 4 } \left[ \frac { 64+8 }{ 3 } \right] \)
\(=\frac { 1 }{ 2 } (30)-\frac { 1 }{ 4 } \left( \frac { 72 }{ 3 } \right) \)
\(=15-\frac { 1 }{ 4 } (24)=15-6\)
= 9 sq.units.

y = 3x² - x is an open upward parabola and meets the X-axis at x = 0, x = \(\frac{1}{3}\) when y = 0
∴ The required area is the combination of 3 pies
\(A=\int _{ 1 }^{ 0 }{ ydx+ } \int _{ 0 }^{ \frac { 1 }{ 3 } }{ -ydy } +\int _{ \frac { 1 }{ 3 } }^{ 1 }{ dy } \) [Since the second area lies below the X-axis]
\(A=\int _{ -1 }^{ 0 }{ ({ 3x }^{ 2 }-x)dx+\int _{ 0 }^{ \frac { 1 }{ 3 } }{ \left( x-3{ x }^{ 2 } \right) dx+ } } \int _{ \frac { 1 }{ 3 } }^{ 1 }{ ({ 3x }^{ 2 }-x)dx } \)
\(={ \left[ \frac { { 3x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } \right] }_{ -1 }^{ 0 }+{ \left[ \frac { { x }^{ 2 } }{ x } -\frac { { 3x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ \frac { 1 }{ 3 } }+{ \left[ \frac { { 3x }^{ 3 } }{ 3 } -\frac { { x }^{ 2 } }{ 2 } \right] }_{ \frac { 1 }{ 3 } }^{ 1 }\\ \)
\(=0-\left( -1-\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 18 } -\frac { 1 }{ 27 } \right) -0+\left( 1-\frac { 1 }{ 2 } \right) -\left( \frac { 1 }{ 27 } -\frac { 1 }{ 18 } \right) \)
\(=\frac { 3 }{ 2 } +\frac { 1 }{ 18 } -\frac { 1 }{ 27 } +\frac { 1 }{ 2 } -\frac { 1 }{ 27 } +\frac { 1 }{ 18 } \)
\(=\left( \frac { 3 }{ 2 } +\frac { 1 }{ 2 } \right) +\left( \frac { 1 }{ 18 } +\frac { 1 }{ 18 } \right) -\left( \frac { 1 }{ 27 } +\frac { 1 }{ 27 } \right) \)
\(=2+\frac { 2 }{ 18 } -\frac { 2 }{ 27 } =2+\frac { 1 }{ 9 } -\frac { 2 }{ 27 } =\frac { 54+3-2 }{ 27 } \)
\(A=\frac { 55 }{ 27 } \) sq.units.

Given C'(x) = 2 + 3e3x
\(\Rightarrow \int { C'(x) } =\int { (2+{ 3e }^{ 3x }) } dx\)
\(\Rightarrow C(x)=2x+\frac { { 3e }^{ 3x } }{ 3 } +k\)
\(\Rightarrow C(x)=2x+{ e }^{ 3x }+k\)
Since the fixed cost is Rs. 500, when x = 0, C = 500
\(\Rightarrow 500=0+{ e }^{ 0 }+k\quad [\because { e }^{ 0 }=1]\)
⇒ 500-1 = k ⇒ = 499
∴ C(x) = 2x + e^3x+499
Average cost dunction \(AC=\frac { C }{ x } \)
\(AC=\frac { 2x+{ e }^{ 3x }+499 }{ x } \)
\(=2+\frac { { e }^{ 3x } }{ x } +\frac { 499 }{ x } \)

Given \({ \eta }_{ d }=\frac { x-5 }{ x } \)

\(\Rightarrow \frac { dx }{ x-5 } =\frac { dp }{ p } \) Integrating both sides,
\(\int { \frac { dx }{ x-5 } } =-\int { \frac { dp }{ p } } \)
log (x - 5)= - log p + log k
log (x - 5) + logp= log k
logp(x - 5) = log k
\(p(x-5)=k\Rightarrow p=\frac { k }{ x-5 } \)
Given that when
p = 2, x = 7
2(7-5) = k ⇒ 2(2) = k
⇒ k = 4
The demand function is
\(p=\frac { 4 }{ x-5 } \)
Revenue, \(R=px=\left( \frac { 4 }{ x-5 } \right) x\)
\(=\frac { 4x }{ x-5 } ,x>5\)

Given' R'(x) = 7 + e^-0.05x
Total revenue from sale of 100 units is
\(R=\int _{ 0 }^{ 100 }{ (7+{ e }^{ -0.05x })dx } \)
\(={ \left[ 7x+\frac { { e }^{ -0.05x } }{ -0.05 } \right] }_{ 0 }^{ 100 }\)
\(=700-\frac { 100 }{ 5 } ({ e }^{ -5 }-{ e }^{ 0 })\left[ 0.05=\frac { 5 }{ 100 } \right] \)
= 700-20(0.0067-1)[∵e⁰=1]
= 700-0.134
= 719.866
Since the revenue is given in thousands,
Total revenue = 719.866 x 1000
= Rs. 7,19,866

Given \(C'(x)=20+\frac { x }{ 20 } \)
\(\Rightarrow \int { C'(x)dx= } \int { \left( 20+\frac { x }{ 20 } \right) dx } \)
\(C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +{ k }_{ 1 }\)
Given when x = 0, C = 200
⇒ k₁ = 200
\(\therefore C(x)=20x+\frac { { x }^{ 2 } }{ 40 } +200\quad ---(1)\)
Also, R'(x) = 30
\(R(x)=\int { 30dx } +{ k }_{ 2 }=30x+{ k }_{ 2 }\)
When x = 0 R = 0 ⇒ K₂ = 0
∴ R(x) = 30x   ----(2)
Profit = Total revenue - total cost
\(=30x-20x-\frac { { x }^{ 2 } }{ 40 } -200\)
\(\therefore P=10x-\frac { { x }^{ 2 } }{ 40 } -200---(3)\)
\(\frac { dp }{ dx } =10-\frac { 2x }{ 40 } =10-\frac { x }{ 20 } \)
\(\frac { dp }{ dx } =0\)
\(\Rightarrow 10-\frac { x }{ 20 } =0\)
\(\Rightarrow 10=\frac { x }{ 20 } \Rightarrow x=200\)
\(\frac { d^{ 2 }p }{ { dx }^{ 2 } } =-\frac { 1 }{ 20 } <0\)
∴ Profit is maximum when x= 200
∴ maximum profit \(P=10(200)-\frac { { (200) }^{ 2 } }{ 40 } -200[From(3)]\)
P = 2000 -1000 - 200
2000 - 1200
P = Rs .800.
Hence, the maximum profit is Rs. 800.

For market equilibrium, P_d = P_s
\({ P }_{ d }=\frac { 16 }{ x+4 } =\frac { x }{ 2 } \Rightarrow 32=x(x+4)\)
⇒ 32-x²+4x
⇒ x²+4x-32 = 0
⇒ (x+8)(x-4) = 0
⇒ x = -8, x = 4
Since x = - 8 is not possible, X_o = 4
\(\therefore { p }_{ 0 }=\frac { 16 }{ 4+4 } =\frac { 16 }{ 8 } =2\)
\(\therefore { p }_{ 0 }{ x }_{ 0 }=2(4)=8\)
Consumer's Surplus \(CS=\int _{ 0 }^{ x0 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(Cs=\int _{ 0 }^{ 4 }{ \frac { 16 }{ x+4 } dx-8 } \)
\(={ \left[ 16log(x+4) \right] }_{ 0 }^{ 4 }-8\)
\(=16\left[ log(4+4)-log(0+4) \right] -8\)
\(=16[log\quad 8-log4]-8\)
\(=16log\left( \frac { 8 }{ 4 } \right) -8\)
CS=(16 log 2-8)units.
Producer's Surplus \(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=8-\int _{ 0 }^{ 4 }{ \frac { x }{ 2 } dx } =8-{ \left[ \frac { { x }^{ 2 } }{ 4 } \right] }_{ 0 }^{ 4 }\)
\(=8-\left[ \frac { { 4 }^{ 2 } }{ 4 } \right] =8-\frac { 16 }{ 4 } =8-4\)
PS = 4units

For market equilibrium, P_d = p_s
⇒ 16-x² = 2x²+4
⇒ 16-4 = 2x²+x²
⇒ 3x² = 12
⇒ x² = 4
⇒ x = 土2
Since x = 2 is not possible x₀ = 2
p₀ = 16 - 2² = 16 - 4 = 12
∴ p₀x₀ = 2 x 12 = 24
Consumer's Surplus CS \(\int _{ 0 }^{ x0 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ (16-{ x }^{ 2 }) } dx-24\)
\(={ \left[ 16x-\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 2 }-24\)
\(=16(2)-\frac { { 2 }^{ 3 } }{ 3 } -24\)
\(=32-\frac { 8 }{ 3 } -24\)
\(=8-\frac { 8 }{ 3 } -24\)
\(=\frac { 16 }{ 3 } \) units
Producer's Surplus \(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=24-\int _{ 0 }^{ 2 }{ (2{ x }^{ 2 }+4)dx } \)
\(=24-{ \left[ \frac { { 2x }^{ 3 } }{ 3 } +4x \right] }_{ 0 }^{ 2 }\)
\(=24-\left[ \frac { 16 }{ 3 } +8 \right] \)
\(=24-\frac { 16 }{ 3 } -8=16-\frac { 16 }{ 3 } \)
\(=\frac { 48-16 }{ 3 } =\frac { 32 }{ 3 } \)units

C(x) = 10 - 6x
⇒ ഽC(x) = ഽ10.6x dx
\(\Rightarrow C(x)=10.6\frac { { x }^{ 2 } }{ 2 } +{ k }_{ 1 }\)
=5.3x2+k1
Given fixed cost is Rs. 50
When x=s 0, C=50 ⇒ k₁ =50
∴ C(x)  5.3x² + 50 ----(1)
Total revenue =number of units sold x price per unit
∴ Pofit =R(x)-C(x)
=5x-(5.3x²+50)
[From (1)&(2)]
P=5x-5.3x²-50

Given that η_d=2
\(\Rightarrow \frac { -p }{ x } .\frac { dx }{ dp } =2\)
\(\Rightarrow \frac { dx }{ x } =-2\frac { dp }{ p } \)
Integrating both sides,
\(\int { \frac { dx }{ x } =-2\int { \frac { dp }{ p } +logk } } \)
⇒ log x+2log p=log k
log x.p²=log k[∵ a log b=log b^a]
⇒ xp²=k
When x=4, p=
4(1)²=k
⇒ k=4
∴ xp²=4
⇒ xp²=\(\frac{4}{x}\)
⇒ \(p-\frac { \sqrt { 4 } }{ x } =\frac { 2 }{ \sqrt { x } } \)
Revenue \(R=px=\frac { 2 }{ \sqrt { x } } .x\)

=2√x
\(R=2\sqrt { x } and\quad p=\frac { 2 }{ \sqrt { x } } \)

Given that MR \(=\frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 }\)
\(\int { MR } =\int { \left( \frac { { e }^{ x } }{ 100 } +x+{ x }^{ 2 } \right) } dx\)
\(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +k\)
When x=0, R=0
\(\Rightarrow 0=\frac { { e }^{ 0 } }{ 100 } +0+0+k\)
\(k=-\frac { 1 }{ 100 } [\because { e }^{ 0 }=1]\)
∴ Revenue \(R=\frac { { e }^{ x } }{ 100 } +\frac { { x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } +\frac { 1 }{ 100 } \)

y = x - 5

The required area lies partially above X-axis and partially below X-axis
∴ Area \(=\int _{ 3 }^{ 5 }{ -ydx } +\int _{ 5 }^{ 7 }{ ydx } \)
\(=\int _{ 3 }^{ 5 }{ (5-x)dx+ } \int _{ 7 }^{ 5 }{ (x-5) } dx\)
\(=\left[ \because \quad y=x-5\Rightarrow -y=5-x \right] \)
\(={ \left[ 5x-\frac { { x }^{ 2 } }{ 2 } \right] }_{ 3 }^{ 5 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -5x \right] }_{ 5 }^{ 7 }\)
\(=\left( 5(5)-\frac { 25 }{ 2 } \right) -\left( 15-\frac { 9 }{ 2 } \right) +\left( \frac { 49 }{ 2 } -35 \right) -\left( \frac { 25 }{ 2 } -25 \right) \)
\(=25-\frac { 25 }{ 2 } -15+\frac { 9 }{ 2 } +\frac { 49 }{ 2 } -35-\frac { 25 }{ 2 } +25\)
\(=(25-15-35+25)+\left( \frac { 25 }{ 2 } +\frac { 9 }{ 2 } +\frac { 49 }{ 2 } -\frac { 25 }{ 2 } \right) \)
\(=0+\left( \frac { -25+9+49-25 }{ 2 } \right) \)
\(=\frac { 58-50 }{ 2 } =\frac { 8 }{ 2 } \)
A = 4 sq.units