Given MC = 8 + 6x
\(C(x)=\int { (8+6x)dx } \) + k₁
= 8x + 3x² + k_{1 ...}(1)
But given when x = 0, C = 0 ⇒ k₁ = 0
ஃ C(x) = 8x + 3x²  ....(2)
Given that MR = 24
R(x) = \(\int { MR } \) dx + k₂
= \(\int { 24 } \) dx + k₂
= \(\int { 24 } \) + k₂
Revenue = 0, when x = 0 ⇒ k₂ = 0
R(x) = 24x ...(3)
Total Profit functions P(x) = R(x) – C(x)
P(x) = 24x − 8x − 3x²
= 16x − 3x²

Since y = x² is symmetric
about Y-axis, the required
Area = \(2\int _{ 0 }^{ 4 }{ x\quad dy } \)
When \(y={ x }^{ 2 }\Rightarrow x=\sqrt { y } \)
∴ Area \(=2\int _{ 0 }^{ 4 }{ \sqrt { y } dy } \)
\(=2\int _{ 0 }^{ 4 }{ { y }^{ \frac { 1 }{ 2 } }dy } \)
\(=2\times \frac { 2 }{ 2 } { \left[ { y }^{ \frac { 3 }{ 2 } } \right] }_{ 0 }^{ 4 }\)
\(=\frac { 4 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } }-{ 0 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 4 }{ 3 } \left[ { 4 }^{ \frac { 3 }{ 2 } } \right] \)
\(=\frac { 4 }{ 3 } \times { ({ 2 }^{ 2 }) }^{ \frac { 3 }{ 2 } }=\frac { 4 }{ 3 } \times { 2 }^{ 3 }\)
\(A=\frac { 4 }{ 3 } \times 8=\frac { 32 }{ 3 } \) sq.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

\(y=\left| x+3 \right| =\begin{cases} x+3\quad if\quad x\ge -3\quad \\ -(x+3)\quad if\quad x<-3 \end{cases}\)
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