Let f(x) = int e^x (x - 1)(x - 2) dx. Then f | vedclass

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(C) Given $f(x) = \int e^x (x - 1)(x - 2) dx$.To find the interval where $f(x)$ decreases,we need to find $f'(x)$ and set $f'(x) < 0$.By the Fundament

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$(1, 2)$

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(C) Given $f(x) = \int e^x (x - 1)(x - 2) dx$.To find the interval where $f(x)$ decreases,we need to find $f'(x)$ and set $f'(x) By the Fundamental Theorem of Calculus,$f'(x) = \frac{d}{dx} \int e^x (x - 1)(x - 2) dx = e^x (x - 1)(x - 2)$.For $f(x)$ to be a decreasing function,we require $f'(x) So,$e^x (x - 1)(x - 2) Since $e^x > 0$ for all real $x$,the inequality simplifies to $(x - 1)(x - 2) The roots of the quadratic $(x - 1)(x - 2) = 0$ are $x = 1$ and $x = 2$.The expression $(x - 1)(x - 2)$ is negative between its roots.Therefore,$f(x)$ decreases in the interval $(1, 2)$.

Let $f(x) = \int e^x (x - 1)(x - 2) dx$. Then $f$ decreases in the interval -

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