If int_0^1 {{e^{{x^2}}}(x - alpha ),dx = 0,} | vedclass

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(C) Given $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0}$.This can be rewritten as $\int_0^1 {x{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} }$.For th

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$0 < \alpha < 1$

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(C) Given $\int_0^1 {{e^{{x^2}}}(x - \alpha )\,dx = 0}$.This can be rewritten as $\int_0^1 {x{e^{{x^2}}}dx = \alpha \int_0^1 {{e^{{x^2}}}dx} }$.For the left side,let $u = x^2$,then $du = 2x \, dx$,so $x \, dx = \frac{1}{2} du$. When $x=0, u=0$ and when $x=1, u=1$.Thus,$\frac{1}{2} \int_0^1 {e^u du} = \alpha \int_0^1 {{e^{{x^2}}}dx}$.$\frac{1}{2} [e^u]_0^1 = \alpha \int_0^1 {{e^{{x^2}}}dx}$.$\frac{1}{2} (e - 1) = \alpha \int_0^1 {{e^{{x^2}}}dx}$.Therefore,$\alpha = \frac{e - 1}{2 \int_0^1 {{e^{{x^2}}}dx}}$.Since $e^x$ is an increasing function on $[0, 1]$,$1 \le e^{x^2} \le e$. Thus,$\int_0^1 1 \, dx Since $e \approx 2.718$,$\frac{e-1}{2} \approx 0.859$. Since $\int_0^1 e^{x^2} \, dx > 1$,$\alpha = \frac{0.859}{ ext{value} > 1} Also,since $e > 1$,$\alpha > 0$. Thus,$0 < \alpha < 1$.

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

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