Question
If $$\int_0^1 {x{e^{{x^2}}}} dx = \lambda \int_0^1 {{e^{{x^2}}}} dx,$$ then :
A.
$$\lambda = 0$$
B.
$$\lambda \, \in \left( {0,\,1} \right)$$
C.
$$\lambda \, \in \left( { - \infty ,\,0} \right)$$
D.
$$\lambda \, \in \left( {1,\,2} \right)$$
Answer :
$$\lambda \, \in \left( {0,\,1} \right)$$
Solution :
$$\eqalign{
& {\text{Here }}0 < x < 1{\text{ and }}{e^{{x^2}}} > 0.\,\,{\text{So, }}0 < x{e^{{x^2}}} < {e^{{x^2}}} \cr
& \therefore \int_0^1 {0\,dx} < \int_0^1 {x{e^{{x^2}}}dx} < \int_0^1 {{e^{{x^2}}}} dx \cr
& \therefore 0 < \lambda \int_0^1 {{e^{{x^2}}}dx} < \int_0^1 {{e^{{x^2}}}dx} \,\,\,\,\,\, \Rightarrow 0 < \lambda < 1 \cr} $$