Definite Integration Question 272
Question: If $ \int_0^{k}{\frac{dx}{2+8x^{2}}}=\frac{\pi }{16}, $ then $ k= $
Options:
A) 1
B) $ \frac{1}{2} $
C) $ \frac{1}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_0^{k}{\frac{1}{2+8x^{2}}dx=\frac{1}{2}\int_0^{k}{\frac{dx}{1+{{(2x)}^{2}}}=\frac{1}{4}\int_0^{2k}{\frac{dt}{1+t^{2}}}}} $
$ =\frac{1}{4}|{{\tan }^{-1}}t|_0^{2k}=\frac{1}{4}{{\tan }^{-1}}2k $ . Comparing it with the given value, we get $ {{\tan }^{-1}}2k=\frac{\pi }{4}\Rightarrow 2k=1\Rightarrow k=\frac{1}{2} $ .
BETA
