Piecewise Functions - 6

Note that both $\int\limits_ {|x|} ^{x^2} \ln t~ dt$ and $\int\limits_ {a} ^{x} e^t \sin t~ dt$ are continuous for all real $x$ . The only possible discontinuity is at $x = a$ .

The function will be continuous at $x =a$ if $\displaystyle\lim_{x \rightarrow a^{-}} = f(a) = \displaystyle\lim_{x \rightarrow a^{+}}$

Note that $\displaystyle\lim_{x \rightarrow a^{+}} = \int\limits_ {a} ^{a} e^t \sin t~ dt = 0$ for all values of $a$ . Also, $\displaystyle\lim_{x \rightarrow a^{-}}$ is always equal to $f(a)$ .

We must now find values of $a$ such that $f(a) = 0$ .

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$f(a) =\int\limits_ {|a|} ^{a^2} \ln t~ dt = 0$ . When $a = -1, 0,$ and $1$ , both limits of the integral are equal, therefore $f(a)$ is zero. It is an exercise to the reader to show that there are no other solutions to the integral.

Finally, we have $a_1 = -1, a_2 = 0, a_3 = 1$

$\displaystyle\prod_{i = 1} ^{n} (a_i -2)^2 = (-3)^2 \times (-2)^2 \times (-1)^2 = \boxed{36}$ $_\square$