Question
What is the area bounded by the curve $$y = 4x - {x^2} - 3$$ and the $$x$$-axis ?
A.
$$\frac{2}{3}{\text{ sq}}{\text{. unit}}$$
B.
$$\frac{4}{3}{\text{ sq}}{\text{. unit}}$$
C.
$$\frac{5}{3}{\text{ sq}}{\text{. unit}}$$
D.
$$\frac{4}{5}{\text{ sq}}{\text{. unit}}$$
Answer :
$$\frac{4}{3}{\text{ sq}}{\text{. unit}}$$
Solution :
Given curve is $$y = 4x - {x^2} - 3$$
Since, area bounded by $$x$$-axis $$\therefore \,y = 0$$
$$\eqalign{
& \Rightarrow 4x - {x^2} - 3 = 0 \cr
& \Rightarrow {x^2} - 4x + 3 = 0 \cr
& \Rightarrow {x^2} - 3x - x + 3 = 0 \cr
& \Rightarrow \left( {x - 3} \right)\left( {x - 1} \right) = 0 \cr
& \Rightarrow x = 1,\,3 \cr} $$
$$\therefore $$ Required area
$$\eqalign{
& = \int_1^3 {\left( {4x - {x^2} - 3} \right)dx} \cr
& = \left. {\frac{{4{x^2}}}{2} - \frac{{{x^3}}}{3} - 3x} \right|_1^3 \cr
& = \left( {\frac{{36}}{2} - \frac{{27}}{3} - 9} \right) - \left( {\frac{4}{2} - \frac{1}{3} - 3} \right) \cr
& = \left( {18 - 9 - 9} \right) - \left( {2 - \frac{{10}}{3}} \right) \cr
& = 0 - \left( {\frac{{ - 4}}{3}} \right) \cr
& = \frac{4}{3}{\text{ sq}}{\text{. unit}} \cr} $$