
What is the value of the definite integral \[\int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx\] ?
A. \[{e^{1000}} - 1\]
B. \[\dfrac{{{e^{1000}} - 1}}{{e - 1}}\]
C. \[1000\left( {e - 1} \right)\]
D. \[\dfrac{{e - 1}}{{1000}}\]
Answer
162.6k+ views
Hint: Here, a definite integral is given. First, simplify the integral by applying the integration rule \[\int\limits_0^{na} {f\left( x \right) dx} = n\int\limits_0^a {f\left( x \right) dx} \]. Then, calculate the value of \[\left[ x \right]\] in the new obtained interval. After that, solve the integral by using the integration formula \[\int {{e^x}} dx = {e^x}\] . In the end, apply the limits and solve the equation to get the required answer.
Formula Used:\[\int\limits_0^{na} {f\left( x \right) dx} = n\int\limits_0^a {f\left( x \right) dx} \]
\[\int {{e^x}} dx = {e^x}\]
Greatest integer function: \[\left[ x \right] = n\], where \[n \le x < n + 1\]
Complete step by step solution:The given definite integral is \[\int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx\].
Let consider,
\[I = \int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx\]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right) dx} = n\int\limits_0^a {f\left( x \right) dx} \] on the right-hand side.
\[I = 1000\int\limits_0^1 {{e^{x - \left[ x \right]}}} dx\]
Now calculate the value of \[\left[ x \right]\] in the interval 0 to 1.
Apply the greatest integer function \[\left[ x \right] = n\], where \[n \le x < n + 1\].
We get,
In the interval \[0 \le x < 1\]
\[\left[ x \right] = 0\]
Thus,
\[I = 1000\int\limits_0^1 {{e^{x - 0}}} dx\]
\[ \Rightarrow I = 1000\int\limits_0^1 {{e^x}} dx\]
Solve the integral by using the integration formula \[\int {{e^x}} dx = {e^x}\].
\[ \Rightarrow I = 1000\left[ {{e^x}} \right]_0^1\]
Apply the upper and lower limits.
\[ \Rightarrow I = 1000\left( {{e^1} - {e^0}} \right)\]
\[ \Rightarrow I = 1000\left( {e - 1} \right)\]
Therefore, \[\int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx = 1000\left( {e - 1} \right)\].
Option ‘C’ is correct
Note: Students get confused and directly use the integration formula \[\int {{e^x}} dx = {e^x}\] to solve the integral and after that they solve for the greatest value function. Because of that, they get the wrong solution.
Formula Used:\[\int\limits_0^{na} {f\left( x \right) dx} = n\int\limits_0^a {f\left( x \right) dx} \]
\[\int {{e^x}} dx = {e^x}\]
Greatest integer function: \[\left[ x \right] = n\], where \[n \le x < n + 1\]
Complete step by step solution:The given definite integral is \[\int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx\].
Let consider,
\[I = \int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx\]
Apply the integration rule \[\int\limits_0^{na} {f\left( x \right) dx} = n\int\limits_0^a {f\left( x \right) dx} \] on the right-hand side.
\[I = 1000\int\limits_0^1 {{e^{x - \left[ x \right]}}} dx\]
Now calculate the value of \[\left[ x \right]\] in the interval 0 to 1.
Apply the greatest integer function \[\left[ x \right] = n\], where \[n \le x < n + 1\].
We get,
In the interval \[0 \le x < 1\]
\[\left[ x \right] = 0\]
Thus,
\[I = 1000\int\limits_0^1 {{e^{x - 0}}} dx\]
\[ \Rightarrow I = 1000\int\limits_0^1 {{e^x}} dx\]
Solve the integral by using the integration formula \[\int {{e^x}} dx = {e^x}\].
\[ \Rightarrow I = 1000\left[ {{e^x}} \right]_0^1\]
Apply the upper and lower limits.
\[ \Rightarrow I = 1000\left( {{e^1} - {e^0}} \right)\]
\[ \Rightarrow I = 1000\left( {e - 1} \right)\]
Therefore, \[\int\limits_0^{1000} {{e^{x - \left[ x \right]}}} dx = 1000\left( {e - 1} \right)\].
Option ‘C’ is correct
Note: Students get confused and directly use the integration formula \[\int {{e^x}} dx = {e^x}\] to solve the integral and after that they solve for the greatest value function. Because of that, they get the wrong solution.
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