Integral Calculus Question 185
Question: $ \int_{{}}^{{}}{\frac{dx}{\sqrt{x}+\sqrt{x-2}}=} $
[MP PET 1990]
Options:
A) $ \frac{1}{3}[{x^{3/2}}-{{(x-2)}^{3/2}}]+c $
B) $ \frac{2}{3}[{x^{3/2}}-{{(x-2)}^{3/2}}]+c $
C) $ \frac{1}{3}[{{(x-2)}^{3/2}}-{x^{3/2}}]+c $
D) $ \frac{2}{3}[{{(x-2)}^{3/2}}-{x^{3/2}}]+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{\frac{dx}{\sqrt{x}+\sqrt{x-2}}=\frac{1}{2}\int_{{}}^{{}}{\frac{x-(x-2)}{\sqrt{x}+\sqrt{x-2}},dx}} $ $ =\frac{1}{2}\int_{{}}^{{}}{(\sqrt{x}-\sqrt{x-2}),dx}=\frac{1}{2}[ \frac{{x^{3/2}}}{3/2}-\frac{{{(x-2)}^{3/2}}}{3/2} ]+c $ $ =\frac{1}{3}{ {x^{3/2}}-{{(x-2)}^{3/2}} }+c. $
BETA
