b) $$ \int_{0}^{\sqrt{2}} x e^{-\left(x^{2} / 2\right)} d x $$

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The value of the integral is $$ 1 - \frac{1}{e} $$. To evaluate the integral $$ I = \int_{0}^{\sqrt{2}} x e^{-\left(x^{2} / 2\right)} \, dx, $$ we can use the substitution method. Let's set $$ u = \frac{x^2}{2}. $$ Then, we differentiate $$u$$ with respect to $$x$$: $$ du = x \, dx \quad \Rightarrow \quad dx = \frac{du}{x} = \frac{du}{\sqrt{2u}}. $$ Next, we need to change the limits of integration. When $$x = 0$$, $$ u = \frac{0^2}{2} = 0, $$ and when $$x = \sqrt{2}$$, $$ u = \frac{(\sqrt{2})^2}{2} = \frac{2}{2} = 1. $$ Now we can rewrite the integral in terms of $$u$$: $$ I = \int_{0}^{1} e^{-u} \, du. $$ The integral of $$e^{-u}$$ is $$ -e^{-u}. $$ Thus, we evaluate: $$ I = \left[-e^{-u}\right]_{0}^{1} = \left[-e^{-1} - (-e^{0})\right] = -\frac{1}{e} + 1 = 1 - \frac{1}{e}. $$ Therefore, the value of the integral is $$ \int_{0}^{\sqrt{2}} x e^{-\left(x^{2} / 2\right)} \, dx = 1 - \frac{1}{e}. $$

b) \( \int_{0}^{\sqrt{2}} x e^{-\left(x^{2} / 2\right)} d x \)

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b) \( \int_{0}^{\sqrt{2}} x e^{-\left(x^{2} / 2\right)} d x \)

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