Given \(R'(x)=\frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \)
\(\Rightarrow \int { R'(x) } =\int { \left( \frac { 3 }{ { x }^{ 2 } } -\frac { 2 }{ x } \right) } dx\)
\(\Rightarrow R(x)=\frac { -3 }{ x } -2log\quad x+k\)
Given R(1) = 6 ⇒ when x = 1, R = 6
\(\Rightarrow 6=\frac { -3 }{ 1 } -2log1+k\)
⇒ 6 + 3 = k [∵ log 1 = 0]
⇒ k = 9 
\(\therefore R(x)=-\frac { 3 }{ x } -2log\quad x+9\)
Demand function \(P=\frac { R }{ x } \)
\(\\ =\frac { 3 }{ { x }^{ 2 } } -\frac { 2log\quad x }{ x } +\frac { 9 }{ x } \)

Given, marginal cost, C' (x) \(\frac{x}{3000}+2.50\)
\(\int { C'(x) } =\int { \left( \frac { x }{ 300 } +2.50 \right) dx } \)
\(C(x)=\frac { { x }^{ 2 } }{ 6000 } +2.50x+k\)
When x = 0, c = 0 ⇒ k = 0
∴ c(x) = \(\frac{x^2}{6000}+2.50x\)
When x = 3000
Cost of production
\(=\frac { { (3000) }^{ 2 } }{ 6000 } +2.50(3000)\)

= 1500 + 7500
= Rs. 9000

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.

Area \(=\int _{ a }^{ b }{ ydx } \)
\(=\int _{ 3 }^{ 9 }{ \frac { 1 }{ x } dx } \)
\(={ [log\quad x] }_{ 3 }^{ 9 }\)
= log9-log3
\(=log\left( \frac { 9 }{ 3 } \right) \)
A = log 3 sq.units.

Area \(=\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ydx } \)
\(=\int _{ -\frac { \pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ cos\quad xdx } \)
\(=2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cos\quad xdx } \) [∵ cos x is an even function]
\(=2{ [sinx] }_{ 0 }^{ \frac { \pi }{ 2 } }\)
\(=2\left[ sin\frac { \pi }{ 2 } -sin0 \right] =2[1-0]\)
A = 2 sq.units.