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

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(B) Given $f(x) = \int {e^x}(x - 1)(x - 2)dx$. By the Fundamental Theorem of Calculus,$f'(x) = \frac{d}{dx} \int {e^x}(x - 1)(x - 2)dx = {e^x}(x - 1)(

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$(1, 2)$

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(B) Given $f(x) = \int {e^x}(x - 1)(x - 2)dx$. 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 the function $f$ to be decreasing,we must have $f'(x) So,${e^x}(x - 1)(x - 2) Since ${e^x} > 0$ for all real $x$,the inequality simplifies to $(x - 1)(x - 2) Using the sign scheme for the quadratic expression $(x - 1)(x - 2)$,the product is negative between the roots $x = 1$ and $x = 2$. Thus,$f$ 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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