question_answer

Find the area of the region bounded by the parabola \[{{x}^{2}}=4y\] and the line \[x=4y-2.\]

Answer:

Given equations of curves are                         \[{{x}^{2}}=4y\]                                 ?(i)             and       \[x=4y-2\]                                 ?(iii) Eq. (i) represents a parabola which is open upward having vertex (0, 0) and Eq. (ii) represents a straight line. On putting the value of 4y from Eq. (i) in Eq.(ii) we get             \[x={{x}^{2}}-2\] \[\Rightarrow \]   \[{{x}^{2}}-x-2=0\] \[\Rightarrow \]\[{{x}^{2}}-2x+x-2=0\] \[\Rightarrow \] \[x(x-2)+1(x-2)=0\] \[\Rightarrow \] \[(x+1)(x-2)=0\] \[\Rightarrow \]   \[x=-\,1,\,\,2\]    When \[x=-\,1,\] then from Eq. (i), \[y=\frac{1}{4}\] and when x = 2, then from Eq. (i), y = 1 \[\therefore \] Points of intersection of given curves are \[\left( -1,\,\,\frac{1}{4} \right)\] and (2, 1). Now, the graph of given curves is as follows:             \[\therefore \] Required area = Area of shaded region BOAB \[=\int_{-1}^{2}{[{{y}_{(line)}}-{{y}_{(parabola)}}]}\,dx=\int_{-1}^{2}{\left[ \left( \frac{x+2}{4} \right)-\frac{{{x}^{2}}}{4} \right]}\,dx\]\[=\frac{1}{4}\int_{-1}^{2}{(x+2-{{x}^{2}})}\,dx=\frac{1}{4}\left[ \frac{{{x}^{2}}}{2}+2x-\frac{{{x}^{3}}}{3} \right]_{-1}^{2}\] \[=\frac{1}{4}\left[ \left( 2+4-\frac{8}{3} \right)-\left( \frac{1}{2}-2+\frac{1}{3} \right) \right]\] \[=\frac{1}{4}\left( 6-\frac{8}{3}+2-\frac{5}{6} \right)\] \[=\frac{1}{4}\left( 8-\frac{8}{3}-\frac{5}{6} \right)\] \[=\frac{1}{4}\left( \frac{48-16-5}{6} \right)=\frac{1}{4}\cdot \frac{27}{6}=\frac{27}{24}=\frac{9}{8}\] Hence, the required area is \[\frac{9}{8}\]sq units.