Question

Find the value of \[\int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} \], where \[\left[ . \right]\] is the greatest integer function.
A. \[31\]
B. \[22\]
C. \[23\]
D. None of these

Answer 1

Hint: Here, a definite integral is given. The function present in the integral is a greatest integer function. First, find the squares present in the given interval and based on that break the interval. Then, calculate the values of the greatest integer function using the interval. After that, split the integrals and solve them using the integration formula. In the end, apply the limits and solve them to get the required answer.



Formula Used:Greatest integer function: \[\left[ x \right] = n\], where \[n \le x < n + 1\]
\[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\], where \[n\] is a number



Complete step by step solution:Given:
The definite integral is \[\int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} \], where \[\left[ . \right]\] is the greatest integer function.

The interval of the given integral is \[\left[ {0,9} \right]\]. The squares present in this interval are \[1,4\] and \[9\].
To split the given integral, find out the different values of the function in that interval.

In the interval \[0 \le x < 1\] :
\[2 \le \sqrt x + 2 < 3\]
Apply the greatest integer function \[\left[ x \right] = n\], where \[n \le x < n + 1\] .
We get, \[\left[ {\sqrt x + 2} \right] = 2\]

In the interval \[1 \le x < 4\] :
\[3 \le \sqrt x + 2 < 4\]
Apply the greatest integer function \[\left[ x \right] = n\], where \[n \le x < n + 1\] .
We get, \[\left[ {\sqrt x + 2} \right] = 3\]

In the interval \[4 \le x < 9\] :
\[4 \le \sqrt x + 2 < 5\]
Apply the greatest integer function \[\left[ x \right] = n\], where \[n \le x < n + 1\] .
We get, \[\left[ {\sqrt x + 2} \right] = 4\]

Now using the above intervals and its values split the given integral.
\[\int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} = \int\limits_0^1 {2dx} + \int\limits_1^4 {3dx} + \int\limits_4^9 {4dx} \]
Solve the right-hand side by using the integration formula \[\int\limits_a^b {ndx} = \left[ {nx} \right]_a^b = n\left( {b - a} \right)\].
We get,
\[\int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} = \left[ {2x} \right]_0^1 + \left[ {3x} \right]_1^4 + \left[ {4x} \right]_4^9\]
Apply the upper and lower limits.
\[\int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} = 2\left( {1 - 0} \right) + 3\left( {4 - 1} \right) + 4\left( {9 - 4} \right)\]
\[ \Rightarrow \int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} = 2\left( 1 \right) + 3\left( 3 \right) + 4\left( 5 \right)\]
\[ \Rightarrow \int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} = 2 + 9 + 20\]
\[ \Rightarrow \int\limits_0^9 {\left[ {\sqrt x + 2} \right]dx} = 31\]



Option ‘A’ is correct



Note: Students often get confused while calculating the value of the greatest integer function of the given function \[\left[ {\sqrt x + 2} \right]\].
To calculate the value, substitute the least value of that interval in the given function and solve it. The obtained value is the value of the function for the corresponding interval.