The value of alpha for which 4 alpha int_{-1} | vedclass

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(C) Given the integral $I = \int_{-1}^{2} e^{-\alpha |x|} dx$. Since $|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$,we split the integral: $I = \i

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$\log_{e} 2$

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(C) Given the integral $I = \int_{-1}^{2} e^{-\alpha |x|} dx$. Since $|x| = -x$ for $x $I = \int_{-1}^{0} e^{\alpha x} dx + \int_{0}^{2} e^{-\alpha x} dx$. Evaluating the integrals: $I = \left[ \frac{e^{\alpha x}}{\alpha} \right]_{-1}^{0} + \left[ \frac{e^{-\alpha x}}{-\alpha} \right]_{0}^{2} = \left( \frac{1}{\alpha} - \frac{e^{-\alpha}}{\alpha} \right) + \left( \frac{1}{\alpha} - \frac{e^{-2\alpha}}{\alpha} \right) = \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha}$. Now,substitute this into the given equation $4\alpha I = 5$: $4\alpha \left( \frac{2 - e^{-\alpha} - e^{-2\alpha}}{\alpha} \right) = 5$. $4(2 - e^{-\alpha} - e^{-2\alpha}) = 5 \Rightarrow 8 - 4e^{-\alpha} - 4e^{-2\alpha} = 5$. $4e^{-2\alpha} + 4e^{-\alpha} - 3 = 0$. Let $t = e^{-\alpha}$. Then $4t^2 + 4t - 3 = 0$. Solving the quadratic equation: $t = \frac{-4 \pm \sqrt{16 - 4(4)(-3)}}{2(4)} = \frac{-4 \pm \sqrt{64}}{8} = \frac{-4 \pm 8}{8}$. Since $t = e^{-\alpha} > 0$,we take $t = \frac{4}{8} = \frac{1}{2}$. $e^{-\alpha} = \frac{1}{2} \Rightarrow e^{\alpha} = 2 \Rightarrow \alpha = \log_{e} 2$.

The value of $\alpha$ for which $4 \alpha \int_{-1}^{2} e^{-\alpha |x|} dx = 5$ is:

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