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.

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 \(AC=\frac { C }{ x } \)
\(AC=\frac { 2x+{ e }^{ 3x }+499 }{ x } \)
\(=2+\frac { { e }^{ 3x } }{ x } +\frac { 499 }{ x } \)

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 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.

\(\frac { EY }{ Ex } =\frac { -7x }{ (1-2x)(2+3x) } \)
\(\frac { 7 }{ (2x-1)(3x+2) } =\frac { A }{ 2x-1 } +\frac { B }{ 3x+2 } \)
7 = A(3x+2)+B(2x-1)
\(Put\ x=\frac { -2 }{ 3 } 7=B\left( \frac { -4 }{ 3 } -1 \right) 7=B\left( \frac { -7 }{ 3 } \right) \)
\(Put\ x=\frac { 1 }{ 2 } 7=A\left( \frac { 3 }{ 2 } +2 \right) \Rightarrow 7=A\left( \frac { 7 }{ 2 } \right) \)
\(\therefore \frac { 7 }{ (2x-1)(3x+2) } =\frac { 2 }{ 2x-1 } -\frac { 3 }{ 3x+2 } \)
Also, it is given that x = 2, when y \(=\frac{3}{8}\)
\(\Rightarrow \frac { x }{ y } \frac { dy }{ dx } =\frac { -7x }{ (1-2x)(2+3x) } \)

\(=\frac { -7dx }{ (1-2x)(2+3x) } \)
\(\Rightarrow \frac { dy }{ y } =\frac { 7dx }{ (2x+1)(3x+2) } \)
\(\int { \frac { dy }{ y } =\int { \frac { 7dx }{ (2x-)(3x+2) } } } \)
\(\int { \frac { dy }{ y } =\int { \left( \frac { 2 }{ 2x-1 } -\frac { 3 }{ 3x+2 } \right) dx } } \)
\(log\quad y=2\int { \frac { 1 }{ 2x-1 } dx-3\int { \frac { dx }{ 3x+2 } } } \)
\(=2\frac { log|2x-1| }{ 2 } -3\frac { log|3x+2| }{ 3 } +logc\)
\(=log\quad y-log\quad c=log\left| \frac { 2x-1 }{ 3x+2 } \right| \)
\(\Rightarrow log\left| \left( \frac { y }{ c } \right) \right| =log\left| \frac { 2x-1 }{ 3x+2 } \right| \)
\(\Rightarrow \frac { y }{ c } =\frac { 2x-1 }{ 3x+2 } y=c\left( \frac { 2x-1 }{ 3x+2 } \right) \) ....(1)
When \(x=2,y=\frac { 3 }{ 8 } \)
\(\Rightarrow \frac { 3 }{ 8 } =c\left( \frac { 4-1 }{ 8 } \right) \)
\(\frac { 3 }{ 8 } =c\left( \frac { 3 }{ 8 } \right) =c=1\)
\(y=\left( \frac { 2x-1 }{ 3x+2 } \right) \)
\(\Rightarrow y= \frac { 2x-1 }{ 3x+2 }\)

Given C(x) = \(\int { C' } (x)dx+\)k₁
= \(\int { \left( 50+\frac { x }{ 50 } \right) } \)dx +k₁
C(x) = 50x + \(\frac{x^2}{100}\)+k₁
When quantity produced is zero, then the fixed cost is 200.
i.e. When x = 0, c = 200
⇒ k₁ = 200
Cost function is C(x) = 50x + \(\frac{x^2}{100}\) + 200        (1)
The Revenue R'(x) = 60
R(x) = \(\int { R' } (x)dx\) + k₂
\(\int { 60 } dx\)+ k₂
= 60x + k₂
When no product is sold, revenue = 0
i.e. When x = 0, R = 0
Revenue R(x) = 60x          (2)
Profit P = Total Revenue – Total cost
= 60x - 50x - \(\frac { { x }^{ 2 } }{ 100 } \) - 200
= 10x - \(\frac { { x }^{ 2 } }{ 100 } \) - 200
\(\frac { dp }{ dx } =10-\frac { x }{ 50 } \)
To get profit maximum, \(\frac { dp }{ dx } \) = 0 ⇒ x = 500
\(\frac { { d }^{ 2 }P }{ { dx }^{ 2 } } =\frac { -1 }{ 50 } <0\)
ஃ Profit is maximum when x = 500 and
Maximum Profit is P = 10(500) - \(\frac { { (500) }^{ 2 } }{ 100 } \) - 200
= 5000 – 2500 – 200
= 2300
Profit = Rs.  2,300.

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

The limits for one arch of the curve y = sin ax When y = 0 ⇒Sin ax = 0
⇒ sin ax = sin 0, sin \(\pi\)
⇒ ax = 0 or ax = \(\pi\)
⇒ x = 0, x = \(\frac{\pi}{a}\)
∴ The limits are from x = 0 to x = \(\frac{\pi}{a}\)
∴ Area =\(\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 0 }^{ a }{ sin\quad ax\quad dx } \)
\(={ \left[ -\frac { cos\quad ax }{ a } \right] }_{ 0 }^{ \frac { \pi }{ a } }\)
\(=-\frac { 1 }{ a } \left[ cos\quad a\times \frac { \pi }{ a } -cos(a)(0) \right] \)
\(=-\frac { 1 }{ a } \left[ cos\quad \pi -cos0 \right] \)
\(=-\frac { 1 }{ a } (-1-1)[\because cos0=1\ cos\pi =-1]\)
\(A=\frac { 2 }{ a } \) 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 marginal cost function = \(\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow MC=\frac { a }{ \sqrt { ax+b } } \Rightarrow \frac { dC }{ dx } =\frac { a }{ \sqrt { ax+b } } \)
\(\Rightarrow dC=\frac { a }{ \sqrt { ax+b } } dx\)
\(\Rightarrow \int { dC } =a\int { \frac { dx }{ \sqrt { ax+b } } } \)
\(\Rightarrow C=a\int { { (ax+b) }^{ \frac { -1 }{ 2 } }dx } \)
\(\Rightarrow C=\not a \frac{(a x+b)^{\frac{-1}{2}+1}}{\left(-\frac{1}{2}+1\right) \not a}+k\)
\(\Rightarrow C=\frac { { (ax+b) }^{ \frac { 1 }{ 2 } } }{ 1 } +k\)
\(\Rightarrow C=2\sqrt { ax+b } +k\) ...(1)
Since the cost of output is zero.
C = 0, when x = 0
\(\therefore (1)\rightarrow 0=2\sqrt { 0+b } +k\)
\(\Rightarrow 0=2\sqrt { b } +k\)
\(\Rightarrow k=-2\sqrt { b } \)
∴(1)becomes
\(C=2\sqrt { ax+b } -2\sqrt { b } \)

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.

Saving Cost S(t) = \(\int _{ 0 }^{ t }{ 20000t } \ dt\)
= 10000 t²
To recoup the total price,
10000 t² = 640000
t² = 64
t = 8
When t = 8 years, one can recoup the price.

Given supply function is P = x + x + 3 and X_o = 4
∴ P_o = 4² + 4 + 3
= 16+ 4 + 3 = 23
∴ p₀x₀ = 23(4) = 92
Producer's surplus
\(PS={ p }_{ 0 }{ x }_{ 0 }-\int _{ 0 }^{ x0 }{ g(x)dx } \)
\(=92-\int _{ 0 }^{ 4 }{ \left( { x }^{ 2 }+x+3 \right) dx } \)
\(=92-{ \left( \frac { { x }^{ 3 } }{ 3 } +\frac { { x }^{ 2 } }{ 2 } +3x \right) }_{ 0 }^{ 4 }\)
\(=92-\left( \frac { { 4 }^{ 3 } }{ 3 } +\frac { { 4 }^{ 2 } }{ 2 } +3(4) \right) \)
\(=92-\left( \frac { 64 }{ 3 } +8+12 \right) \)
\(=92-\frac { 64 }{ 3 } -20=72-\frac { 64 }{ 3 } \)
\(=\frac { 216-64 }{ 3 } \)
\(PS=\frac { 152 }{ 3 } \) units

Given demand function is p = 25 - x - X²
and p₀ = 19
⇒ 19 = 25-x-x²
⇒ x² + x - 6 = 0
⇒ (x + 3) (x - 2) = 0
⇒ x = -3 or x = 2
Since x cannot be negative x_o = 2
p_o x_o = 19(2) = 38
\(CS=\int _{ 0 }^{ 2 }{ f(x)dx-{ p }_{ 0 }{ x }_{ 0 } } \)
\(=\int _{ 0 }^{ 2 }{ \left( 25-x-{ x }^{ 2 } \right) dx-38 } \)
\(={ \left( 25x-\frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right) }_{ 0 }^{ 2 }-38\)
\(=25(2)-\frac { 4 }{ 2 } -\frac { 8 }{ 3 } -38\)
\(=50-2-\frac { 8 }{ 3 } -38\)
\(=10-\frac { 8 }{ 3 } =\frac { 30-8 }{ 3 } \)
\(CS=\frac { 22 }{ 3 } \) units