Given MR = 9 - 6x² + 2x
⇒ഽMR=ഽ(9 - 6x² + 2x)sx
\(\Rightarrow R=9x-\frac { { 6x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 2 } }{ 2 } +k\)
⇒ R = 9x - 2x³ + x² + k
When x = 0, R = 0 ⇒ k = 0
∴ R = 9x - 2x³ + x²

Given MC = 3x² - 2x + 8
⇒ ഽMC = ഽ(3x² - 2x + 8)dx
\(\Rightarrow C=\frac { { 3x }^{ 3 } }{ 3 } -\frac { { 2x }^{ 2 } }{ 2 } +8x+k\)
⇒ C = x³ - x² + 8x + k
Since there is no fixed cost,
when x = 0, C = 0 ⇒ k = 0
∴ C = x³ - x² + 8x

Given curve is y = 4x - X²
The limits are from x = 0 to x = 3
∴ Area \(\int _{ 0 }^{ 3 }{ y } dx=\int _{ 0 }^{ 3 }{ (4x-{ x }^{ 2 }) } dx\)
\({ \left[ \frac { 4{ x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 3 } \right] }_{ 0 }^{ 3 }=2(9)-\frac { 27 }{ 3 } \)
=18 - 9
Area = 9 sq.units

The Y-axis is the ordinate at x = 0.
The area bounded by the ordinates at x = 0, x = 2 and the given curve is
\(A=\int _{ a }^{ b }{ y } dx\)
\(=\int _{ 0 }^{ 2 }{ ({ 4x }^{ 2 }-8x+6)dx } \)
\(={ \left[ \frac { { 4x }^{ 3 } }{ 3 } -\frac { 8{ x }^{ 2 } }{ 2 } +6x \right] }_{ 0 }^{ 2 }\)
\(=4\left( \frac { 8 }{ 3 } \right) -4(4)+6(2)\)
\(\frac { 32 }{ 3 } -6+12=\frac { 32 }{ 3 } -4\)
\(=\frac { 32-12 }{ 3 } =\frac { 20 }{ 3 } \)
∴ Area \(=\frac { 20 }{ 3 } \)sq.units

Area under the curve is
\(A=\int _{ c }^{ d }{ xdy } =\int _{ 2 }^{ 4 }{ \sqrt { 4y } dy } \)
\(=2\int _{ 2 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } =2{ \left( \frac { { y }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } \right) }_{ 2 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left( { y }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( { 4 }^{ \frac { 3 }{ 2 } }-{ 2 }^{ \frac { 3 }{ 2 } } \right) \)
\(=\frac { 4 }{ 3 } \left( 4\sqrt { 4 } -2\sqrt { 2 } \right) \)
\(=\frac { 4 }{ 3 } (8-2\sqrt { 2 } )\)sq.units