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If $I=\int_{0}^{2} e^{x^{4}}(x-a) d x=0$, then $\alpha$ lies in the interval
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The correct answer is:
(0,2)
Given, $I=\int_{0}^{2} e^{x^{4}}(x-\alpha) d x=0$
$\Rightarrow \quad \int_{0}^{2} e^{x^{4}} x d x=\int_{0}^{2} e^{x^{4}} \alpha d x$
Here, we see that $\int_{0}^{2} e^{x^{4}} x d x$ gives us an area between two curves $e^{x^{4}}$ and $x$ from $x=0$ to $x=2$
Similarly. $\int_{0}^{2} e^{x^{4}} \alpha d x$ gives us a same area, so $a$ should lies between 0 to $2, i . e ., \alpha \in(0,2)$
$\Rightarrow \quad \int_{0}^{2} e^{x^{4}} x d x=\int_{0}^{2} e^{x^{4}} \alpha d x$
Here, we see that $\int_{0}^{2} e^{x^{4}} x d x$ gives us an area between two curves $e^{x^{4}}$ and $x$ from $x=0$ to $x=2$
Similarly. $\int_{0}^{2} e^{x^{4}} \alpha d x$ gives us a same area, so $a$ should lies between 0 to $2, i . e ., \alpha \in(0,2)$
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