Definite Integration Question 2
Question: If $ \int_0^{1}{{e^{x^{2}}}(x-\alpha )dx=0,} $ then
[MNR 1994; Pb. CET 2001; UPSEAT 2000]
Options:
A) $ 1<\alpha <2 $
B) $ \alpha <0 $
C) $ 0<\alpha <1 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{1}{{e^{x^{2}}}}(x-\alpha )dx=0 $
therefore $ \frac{1}{2}\int_0^{1}{2x.{e^{x^{2}}}dx=\alpha \int_0^{1}{{e^{x^{2}}}dx}} $
therefore $ \frac{1}{2}|{e^{x^{2}}}|_0^{1}=\alpha \int_0^{1}{{e^{x^{2}}}dx} $
therefore $ \frac{1}{2}(e-1)=\alpha \int_0^{1}{{e^{x^{2}}}dx} $
therefore $ \alpha =\frac{\frac{1}{2}(e-1)}{\int_0^{1}{{e^{x^{2}}}dx}}>0 $ and $ \alpha <1 $ . So, $ 0<\alpha <1 $ .
BETA
