If _0^1 e^ x^2 (x - ) ,dx = 0, then - Vidyadip

MCQ

If $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0,} $ then

A
$1 < \alpha < 2$
B
$\alpha < 0$
✓
$0 < \alpha < 1$
D
None of these

✓

Answer

Correct option: C.

$0 < \alpha < 1$

c
(c) $\int_0^1 {{e^{{x^2}}}} (x - \alpha )\,dx = 0$

==> $\frac{1}{2}\int_0^1 {2x.{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} } $

==> $\frac{1}{2}|{e^{{x^2}}}|_0^1 = \alpha \int_0^1 {{e^{{x^2}}}dx} $

==> $\frac{1}{2}(e - 1) = \alpha \,\int_0^1 {{e^{{x^2}}}dx} $

==> $\alpha = \frac{{\frac{1}{2}(e - 1)}}{{\int_0^1 {{e^{{x^2}}}dx} }} > 0$ and $\alpha < 1$.

So, $0 < \alpha < 1$.

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