Definite Integration Question 259
Question: $ \int_0^{1}{\frac{dx}{{{[ax+b(1-x)]}^{2}}}}= $
[SCRA 1986]
Options:
A) $ \frac{a}{b} $
B) $ \frac{b}{a} $
C) $ ab $
D) $ \frac{1}{ab} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ I=\int_0^{1}{\frac{dx}{{{[(a-b)x+b]}^{2}}}} $
Put $ t=(a-b)x+b\Rightarrow dt=(a-b)dx $
As $ x=1\Rightarrow t=a $ and $ x=0\Rightarrow t=b $ , then $ I=\frac{1}{a-b}\int_b^{a}{\frac{1}{t^{2}}}dt=\frac{1}{(a-b)}[ -\frac{1}{t} ]_b^{a}=\frac{1}{(a-b)}( \frac{a-b}{ab} )=\frac{1}{ab} $ .
BETA
