A) 1 : 1 : 1
B) 2 : 1 : 2
C) 1 : 2 : 3
D) 1 : 2 : 1
Correct Answer: A
Solution :
Now, \[{{S}_{1}}={{S}_{3}}=\int_{0}^{4}{\frac{{{x}^{2}}}{4}}dx=\frac{1}{4}\left[ \frac{{{x}^{3}}}{3} \right]_{0}^{4}\] \[=\frac{1}{12}\times 64=\frac{16}{3}\]sq units \[\therefore \]\[{{S}_{2}}+{{S}_{3}}=\int_{0}^{4}{\sqrt{4x}}dx\] \[=2\,\,\left[ \frac{{{x}^{3/2}}}{3/2} \right]_{0}^{4}\,=\frac{4}{3}\times 8=\,\frac{32}{3}\,sq\] units \[\Rightarrow \]\[{{S}_{2}}=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}\]sq units \[\therefore \]\[{{S}_{1}}:{{S}_{2}}:{{S}_{3}}=\frac{16}{3}:\frac{6}{3}:\frac{16}{3}=1:1:1\]You need to login to perform this action.
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