Definite Integration Question 193
Question: The area bounded by the curves $ x^{2}+y^{2}=25, $
$ 4y=| 4-x^{2} | $ and $ x=0 $ , above x-axis is
Options:
A) $ 2+\frac{25}{2}{{\sin }^{-1}}\frac{4}{5} $
B) $ 2+\frac{25}{4}{{\sin }^{-1}}\frac{4}{5} $
C) $ 2+\frac{25}{2}{{\sin }^{-1}}\frac{1}{5} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The required area $ =\int_0^{4}{\sqrt{25-x^{2}}dx-\int_0^{2}{\frac{4-x^{2}}{4}dx-\int_2^{4}{\frac{x^{2}-4}{4}dx}}} $
$ =[ \frac{x}{2}\sqrt{25-x^{2}}+\frac{25}{2}{{\sin }^{-1}}\frac{x}{5} ]_0^{4} $
$ =-\frac{1}{4}[ 4x-\frac{x^{3}}{3} ]_0^{2}-\frac{1}{4}[ \frac{x^{3}}{3}-4x ]_2^{4}=2+\frac{25}{2}{{\sin }^{-1}}\frac{4}{5}. $
BETA
