
If \[\left[ x \right]\] denotes the greatest integer less than or equal to \[x\], then what is the value of the integral \[\int\limits_0^2 {{x^2}} \left[ x \right]dx\]?
A. \[\dfrac{5}{3}\]
B. \[\dfrac{7}{3}\]
C. \[\dfrac{8}{3}\]
D. \[\dfrac{4}{3}\]
Answer
164.7k+ views
Hint: Here, a definite integral is given. First, simplify the given integral by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]. Then, check the given condition of the greatest integer function and substitute the values in the integral. After that, solve the integral by using the integration formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]. In the end, apply the upper and lower limits and solve it to get the required answer.
Formula Used: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
Greatest integer function: \[\left[ x \right] = n\], where \[n \le x < n + 1\]
Complete step by step solution:Given:
\[\left[ x \right]\] denotes the greatest integer less than or equal to \[x\]
And the definite integral is \[\int\limits_0^2 {{x^2}} \left[ x \right]dx\]
Let consider,
\[I = \int\limits_0^2 {{x^2}} \left[ x \right]dx\]
Simplify the integral by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\].
\[I = \int\limits_0^1 {{x^2}} \left[ x \right]dx + \int\limits_1^2 {{x^2}} \left[ x \right]dx\]
Now apply the condition of the greatest integer function\[\left[ x \right] = n\], where \[n \le x < n + 1\].
We get,
\[I = \int\limits_0^1 {{x^2}} \left( 0 \right)dx + \int\limits_1^2 {{x^2}} \left( 1 \right)dx\]
\[ \Rightarrow I = 0 + \int\limits_1^2 {{x^2}} dx\]
\[ \Rightarrow I = \int\limits_1^2 {{x^2}} dx\]
Solve the integral by using the integration formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[ \Rightarrow I = \left[ {\dfrac{{{x^3}}}{3}} \right]_1^2\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\dfrac{{{2^3}}}{3} - \dfrac{{{1^3}}}{3}} \right]\]
\[ \Rightarrow I = \left[ {\dfrac{8}{3} - \dfrac{1}{3}} \right]\]
\[ \Rightarrow I = \dfrac{7}{3}\]
Therefore, \[\int\limits_0^2 {{x^2}} \left[ x \right]dx = \dfrac{7}{3}\].
Option ‘B’ is correct
Note: Students get confused about the concept of the greatest integer function. It happens because of the word greatest.
The greater integer function is a function that gives the output of the greatest integer that will be less than the input or lesser than the input.
Formula Used: \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\]
\[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\]
Greatest integer function: \[\left[ x \right] = n\], where \[n \le x < n + 1\]
Complete step by step solution:Given:
\[\left[ x \right]\] denotes the greatest integer less than or equal to \[x\]
And the definite integral is \[\int\limits_0^2 {{x^2}} \left[ x \right]dx\]
Let consider,
\[I = \int\limits_0^2 {{x^2}} \left[ x \right]dx\]
Simplify the integral by applying the integration rule \[\int\limits_a^b {f\left( x \right)} dx = \int\limits_a^c {f\left( x \right)} dx + \int\limits_c^b {f\left( x \right)} dx\].
\[I = \int\limits_0^1 {{x^2}} \left[ x \right]dx + \int\limits_1^2 {{x^2}} \left[ x \right]dx\]
Now apply the condition of the greatest integer function\[\left[ x \right] = n\], where \[n \le x < n + 1\].
We get,
\[I = \int\limits_0^1 {{x^2}} \left( 0 \right)dx + \int\limits_1^2 {{x^2}} \left( 1 \right)dx\]
\[ \Rightarrow I = 0 + \int\limits_1^2 {{x^2}} dx\]
\[ \Rightarrow I = \int\limits_1^2 {{x^2}} dx\]
Solve the integral by using the integration formula \[\int\limits_a^b {{x^n}dx = \left[ {\dfrac{{{x^{n + 1}}}}{{n + 1}}} \right]} _a^b\].
\[ \Rightarrow I = \left[ {\dfrac{{{x^3}}}{3}} \right]_1^2\]
Apply the upper and lower limits.
\[ \Rightarrow I = \left[ {\dfrac{{{2^3}}}{3} - \dfrac{{{1^3}}}{3}} \right]\]
\[ \Rightarrow I = \left[ {\dfrac{8}{3} - \dfrac{1}{3}} \right]\]
\[ \Rightarrow I = \dfrac{7}{3}\]
Therefore, \[\int\limits_0^2 {{x^2}} \left[ x \right]dx = \dfrac{7}{3}\].
Option ‘B’ is correct
Note: Students get confused about the concept of the greatest integer function. It happens because of the word greatest.
The greater integer function is a function that gives the output of the greatest integer that will be less than the input or lesser than the input.
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