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Question: For real x with \( -10\le x\le 10 \) define \( f\left( x \right)=\int\limits_{-10}^{x}{2\left[ t \ri...

For real x with −10≤x≤10-10\le x\le 10 define f(x)=∫−10x2[t]dtf\left( x \right)=\int\limits_{-10}^{x}{2\left[ t \right]dt} , where for a real number r we denote [r]\left[ r \right] the largest integer less than or equal to r. The number of points of discontinuity of f in the interval (−10,10)\left( -10,10 \right) is
(a) 0
(b) 10
(c) 18
(d) 19

Explanation

Solution

Hint:We have to find the number of points where discontinuity lies. The point of discontinuity is the point where the continuity of the function breaks. We have given the greatest integer function which will show discontinuity at the integral points. So, the points of discontinuity are the integral values between the given interval i.e. (−10,10)\left( -10,10 \right)

Complete step-by-step answer:
The function given in the above problem is:
f(x)=∫−10x2[t]dtf\left( x \right)=\int\limits_{-10}^{x}{2\left[ t \right]dt}
The value of x in the above integral is real and lies between −10≤x≤10-10\le x\le 10 so we are going to substitute this interval of x in the above integration. The substitution of the value of x is done as follows:
f(x)=∫−10−92(−10)dt+∫−9−82(−9)dt+.......+∫9102(9)dtf\left( x \right)=\int\limits_{-10}^{-9}{2\left( -10 \right)dt+\int\limits_{-9}^{-8}{2\left( -9 \right)dt}+.......+\int\limits_{9}^{10}{2\left( 9 \right)dt}}
In the above integration we have split the interval of x from -10 to -9 then -9 to -8 and so on till 9 to 10.
We know that the value of the largest integer function [t]\left[ t \right] when the value of t is between 0 and 1 is 0. Or in general, we can say that when the value of t lies between n−1&nn-1\And n then the value of [t]\left[ t \right] is n−1n-1 where n is any integer. Using this property of largest integer, we have solved the above integral as when limits of integral lie between -10 and 9 then the value of [t]\left[ t \right] is -10. Similarly, we have done for all the limits of the integral.
f(x)=∫−10−92(−10)dt+∫−9−82(−9)dt+.......+∫9102(9)dtf\left( x \right)=\int\limits_{-10}^{-9}{2\left( -10 \right)dt+\int\limits_{-9}^{-8}{2\left( -9 \right)dt}+.......+\int\limits_{9}^{10}{2\left( 9 \right)dt}}
In the above integral as you can see that in the limits of -10 and -9 the integration gives different results from the limits of -9 and -8 so we can see that at the point -9 the integral value is different which shows that the point -9 is the point of discontinuity. Similarly at all the integral points from -9 up till 10 shows the discontinuity of the function. From this justification of the discontinuity of the points, the number of points in the interval (−10,10)\left( -10,10 \right) where the function is going discontinuous are:
-9, -8, -7, -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
So, the total number of points of the discontinuity is 19.
Hence, the correct option is (d).
Note: The points of mistake that could happen in solving the above problem is that you would have wrongly counted the number of discontinuity points. In the integration from -10 to -9 the value of [t]\left[ t \right] is -10 so you might have tempted to consider -10 as point of discontinuity also which is wrong because we are asked to find the points of discontinuity between (−10,10)\left( -10,10 \right) and this interval is an open interval meaning should not consider the values -10 and 10.