Definite Integration Question 160
Question: What is the area enclosed between the curves $ y^{2}=12x $ and the lines $ x=0 $ and $ y=6 $ -
Options:
A) 2 sq. unit
B) 4 sq. unit
C) 6 sq. unit
D) 8 sq. unit
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Equation of given curve is $ y^{2}=12x $
At $ y=6,36=12x\Rightarrow x=3 $
$ \therefore $ Required area $ =\int_0^{3}{(y_1-y_2)}dx $ where $ y_1 $ represents line and $ y_2 $ represents the curve.
$ =\int_0^{3}{( 6-\sqrt{12x} )dx=[ 6x ]_0^{3}-\sqrt{12}[ \frac{2{x^{3/2}}}{3} ]_0^{3}} $
$ =[ 6\times 3 ]-\frac{\sqrt{12}\times 2\times \sqrt{27}}{3}=18-12=6 $ sq. unit
BETA
