If I = int_{0}^{2} e^{x^{4}}(x - alpha) dx = | vedclass

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Question

(A) Given the integral equation: $\int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$. This can be rewritten as: $\int_{0}^{2} x e^{x^{4}} dx = \alpha \int_{0}

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$(0, 2)$

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(A) Given the integral equation: $\int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$. This can be rewritten as: $\int_{0}^{2} x e^{x^{4}} dx = \alpha \int_{0}^{2} e^{x^{4}} dx$. Thus, $\alpha = \frac{\int_{0}^{2} x e^{x^{4}} dx}{\int_{0}^{2} e^{x^{4}} dx}$. Let $f(x) = e^{x^{4}}$. Since $f(x) > 0$ for all $x \in [0, 2]$, the expression for $\alpha$ represents a weighted average of the values of $x$ in the interval $[0, 2]$ with respect to the weight function $f(x)$. Since $x$ ranges from $0$ to $2$, the weighted average $\alpha$ must also lie strictly between the minimum and maximum values of $x$ in the interval $[0, 2]$. Therefore, $0 < \alpha < 2$, which means $\alpha \in (0, 2)$.

If $I = \int_{0}^{2} e^{x^{4}}(x - \alpha) dx = 0$, then $\alpha$ lies in the interval

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