If beta + 2 intlimits_0^1 {{x^2},{e^{ - {x^2} | vedclass

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(A) Given the equation: $\beta + 2 \int\limits_0^1 x^2 e^{-x^2} dx = \int\limits_0^1 e^{-x^2} dx$. We use integration by parts for the integral $\int\

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$e^{-1}$

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(A) Given the equation: $\beta + 2 \int\limits_0^1 x^2 e^{-x^2} dx = \int\limits_0^1 e^{-x^2} dx$. We use integration by parts for the integral $\int\limits_0^1 x(2x e^{-x^2}) dx$. Let $u = x$ and $dv = 2x e^{-x^2} dx$. Then $du = dx$ and $v = -e^{-x^2}$. Using the formula $\int u dv = uv - \int v du$: $\int\limits_0^1 x(2x e^{-x^2}) dx = [ -x e^{-x^2} ]_0^1 - \int\limits_0^1 (-e^{-x^2}) dx$. Substituting this into the original equation: $\beta + [ -x e^{-x^2} ]_0^1 + \int\limits_0^1 e^{-x^2} dx = \int\limits_0^1 e^{-x^2} dx$. Subtracting $\int\limits_0^1 e^{-x^2} dx$ from both sides: $\beta + [ -x e^{-x^2} ]_0^1 = 0$. $\beta - (1 \cdot e^{-1} - 0 \cdot e^0) = 0$. $\beta - e^{-1} = 0$. $\beta = e^{-1} = \frac{1}{e}$.

If $\beta + 2 \int\limits_0^1 {{x^2}\,{e^{ - {x^2}}}}\, dx = \int\limits_0^1 {{e^{ - {x^2}}}}\, dx$,then the value of $\beta$ is

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