A blend of Graphs and limits!

The shaded region in the above graph represents the given equation . And it sums up to 18. So A=18.

The asymptote is given as $y=\frac { x }{ 2 } +1$ .
We know that if $\lim _{ x\rightarrow \infty }{ \frac { f\left( x \right) }{ x } =m }$ , and $\lim _{ x\rightarrow \infty }{ \left( f\left( x \right) -mx \right) } =c$ , then equation of the asymptote is given as $y=mx+c$ . Comparing this with the given equation of asymptote we have $\lim _{ x\rightarrow \infty }{ \frac { f\left( x \right) }{ x } } =\quad \frac { 1 }{ 2 }$ and $\lim _{ x\rightarrow \infty }{ \left( f\left( x \right) -\frac { x }{ 2 } \right) } =1$ .

$\therefore \quad \lim _{ x\rightarrow \infty }{ { \left[ \frac { { \left[ \left( 2f\left( x \right) +1 \right) \left( 2f\left( x \right) +3 \right) \left( 2f\left( x \right) -5 \right) \left( 2f\left( x \right) -3 \right) \left( 2f\left( x \right) -1 \right) \right] }^{ \frac { 1 }{ 5 } }-f\left( x \right) }{ f\left( x \right) } \right] }^{ x\left( 2f\left( x \right) -x \right) } } \\ We\quad can\quad observe\quad that\quad the\quad limit\quad is\quad of\quad the\quad form\quad { 1 }^{ \infty }\\ =\quad \lim _{ x\rightarrow \infty }{ { \left[ 1+\frac { { \left[ \left( 2f\left( x \right) +1 \right) \left( 2f\left( x \right) +3 \right) \left( 2f\left( x \right) -5 \right) \left( 2f\left( x \right) -3 \right) \left( 2f\left( x \right) -1 \right) \right] }^{ \frac { 1 }{ 5 } }-2f\left( x \right) }{ f\left( x \right) } \right] }^{ x\left( 2f\left( x \right) -x \right) } } \\ =\quad { e }^{ \lim _{ x\rightarrow \infty }{ \left[ \frac { { \left[ \left( 2f\left( x \right) +1 \right) \left( 2f\left( x \right) +3 \right) \left( 2f\left( x \right) -5 \right) \left( 2f\left( x \right) -3 \right) \left( 2f\left( x \right) -1 \right) \right] }^{ \frac { 1 }{ 5 } }-2f\left( x \right) }{ f\left( x \right) } \right] \left( x\left( 2f\left( x \right) -x \right) \right) } }\\ =\quad { e }^{ \lim _{ x\rightarrow \infty }{ \left[ \frac { { \left[ 32{ \left( f\left( x \right) \right) }^{ 5 }-80{ \left( f\left( x \right) \right) }^{ 4 }-80{ \left( f\left( x \right) \right) }^{ 3 }+200{ \left( f\left( x \right) \right) }^{ 2 }+18f\left( x \right) -45 \right] }^{ \frac { 1 }{ 5 } }-{ \left( 32{ \left( f\left( x \right) \right) }^{ 5 } \right) }^{ \frac { 1 }{ 5 } } }{ f\left( x \right) } \right] \left( x\left( 2f\left( x \right) -x \right) \right) } }\\ If\quad we\quad consider\quad t=32{ \left( f\left( x \right) \right) }^{ 5 }-80{ \left( f\left( x \right) \right) }^{ 4 }-80{ \left( f\left( x \right) \right) }^{ 3 }+200{ \left( f\left( x \right) \right) }^{ 2 }+18f\left( x \right) -45\quad and\quad u=32{ \left( f\left( x \right) \right) }^{ 5 }\quad ,\\ { t }^{ \frac { 1 }{ 5 } }-{ u }^{ \frac { 1 }{ 5 } }=\quad \frac { t-u }{ { t }^{ \frac { 4 }{ 5 } }+{ t }^{ \frac { 3 }{ 5 } }{ u }^{ \frac { 1 }{ 5 } }+t^{ \frac { 2 }{ 5 } }{ u }^{ \frac { 2 }{ 5 } }+{ t }^{ \frac { 1 }{ 5 } }{ u }^{ \frac { 3 }{ 5 } }+{ u }^{ \frac { 4 }{ 5 } } } \\ Solving\quad as\quad above\quad ,\quad \\ Limit\quad =\quad { e }^{ \lim _{ x\rightarrow \infty }{ \left[ \frac { -80{ \left( f\left( x \right) \right) }^{ 4 }-80{ \left( f\left( x \right) \right) }^{ 3 }+200{ \left( f\left( x \right) \right) }^{ 2 }+18f\left( x \right) }{ f\left( x \right) \left( 80{ \left( f\left( x \right) \right) }^{ 4 }+\quad terms\quad having\quad lower\quad powers\quad of\quad f\left( x \right) \right) } \right] \left( x\left( 2f\left( x \right) -x \right) \right) } }\quad \\ Now\quad ,\quad by\quad putting\quad \lim _{ x\rightarrow \infty }{ \frac { x }{ f\left( x \right) } =2\quad and\quad \quad \lim _{ x\rightarrow \infty }{ \left( 2f\left( x \right) -x \right) } } =2\quad ,\\ we\quad get\quad the\quad limit\quad as\quad { e }^{ -4 }.\quad So\quad L=-4.\\ \therefore \quad \left| A\times L \right| =\left| 18\times \left( -4 \right) \right| =72$