Integral Calculus Question 464
Question: $ \int_{{}}^{{}}{\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}}= $
[AISSE 1989]
Options:
A) $ \frac{2}{3(b-a)}[{{(x+a)}^{3/2}}-{{(x+b)}^{3/2}}]+c $
B) $ \frac{2}{3(a-b)}[{{(x+a)}^{3/2}}-{{(x+b)}^{3/2}}]+c $
C) $ \frac{2}{3(a-b)}[{{(x+a)}^{3/2}}+{{(x+b)}^{3/2}}]+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}=\int_{{}}^{{}}{\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)},dx}} $
$ =\frac{1}{(a-b)}\int_{{}}^{{}}{{{(x+a)}^{1/2}}dx}-\frac{1}{(a-b)}\int_{{}}^{{}}{{{(x+b)}^{1/2}}dx} $
$ =\frac{2}{3(a-b)}[{{(x+a)}^{3/2}}-{{(x+b)}^{3/2}}]+c. $
BETA
