Definite Integration Question 310
Question: Area bounded by the curve $ x^{2}=4y $ and the straight line $ x=4y-2 $ is
[SCRA 1986; IIT 1981; Pb. CET 2003]
Options:
A) $ \frac{8}{9} $ sq. unit
B) $ \frac{9}{8} $ sq. unit
C) $ \frac{4}{3} $ sq. unit
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Solving the equations $ x^{2}=4y $ and $ x=4y-2 $ simultaneously. The points of intersection of the parabola and the line are $ A(2,1) $ and $ B( -1,\frac{1}{4} ) $ . The required area = shaded area $ =[ \int _{-1}^{2}{yd{x _{(forx=4y-2)}}} ]-[ \int _{-1}^{2}{yd{x _{(forx^{2}=4y)}}} ] $
$ =\int _{-1}^{2}{\frac{1}{4}(x+2)dx-\int _{-1}^{2}{\frac{1}{4}x^{2}dx}} $
$ =\frac{1}{4}[ \frac{x^{2}}{2}+2x ] _{-1}^{2}-\frac{1}{4}[ \frac{x^{3}}{3} ] _{-1}^{2}=\frac{9}{8} $ sq. unit.
BETA
