For real x with -10 leq x leq 10,define f(x) | vedclass

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(A) The function $f(x) = \int_{-10}^x 2^{[t]} dt$ is an integral of a step function. By the Fundamental Theorem of Calculus,if $g(t) = 2^{[t]}$,then $

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$0$

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(A) The function $f(x) = \int_{-10}^x 2^{[t]} dt$ is an integral of a step function. By the Fundamental Theorem of Calculus,if $g(t) = 2^{[t]}$,then $f(x) = \int_{-10}^x g(t) dt$ is continuous everywhere if the integrand $g(t)$ is integrable. Since $g(t) = 2^{[t]}$ is a piecewise constant function,it is continuous except at integers. However,the integral of a piecewise continuous function is always continuous. Specifically,for any integer $n \in (-10, 10)$,the left-hand limit is $\lim_{x 	o n^-} f(x) = \int_{-10}^n 2^{[t]} dt$ and the right-hand limit is $\lim_{x 	o n^+} f(x) = \int_{-10}^n 2^{[t]} dt + \lim_{x 	o n^+} \int_n^x 2^n dt = \int_{-10}^n 2^{[t]} dt + 0 = f(n)$. Since the left-hand limit,right-hand limit,and the value of the function are equal at every point $x \in (-10, 10)$,the function $f(x)$ is continuous everywhere in the interval. Therefore,the number of points of discontinuity is $0$.

For real $x$ with $-10 \leq x \leq 10$,define $f(x) = \int_{-10}^x 2^{[t]} dt$,where for a real number $r$,we denote by $[r]$ the greatest integer less than or equal to $r$. The number of points of discontinuity of $f$ in the interval $(-10, 10)$ is

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