
If \[f\left( x \right) = f\left( {2 - x} \right)\], then find the value of \[\int_{0.5}^{1.5} {xf\left( x \right)dx} \].
A. \[\int_0^1 {f\left( x \right)dx} \]
B. \[\int_{0.5}^{1.5} {f\left( x \right)dx} \]
C. \[2\int_{0.5}^{1.5} {f\left( x \right)dx} \]
D. 0
Answer
162.9k+ views
Hint: The given integration is a definite integral. Thus first we will simplify the integration by using the property of the definite integral. To simplify we will apply the property \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]. Then simplify it to get the required solution.
Formula Used:Definite integral property:
\[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx
Complete step by step solution:Given definite integral is
\[I = \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
Now applying the property of definite integral \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]:
\[ \Rightarrow I = \int_{0.5}^{1.5} {\left( {1.5 + 0.5 - x} \right)f\left( {1.5 + 0.5 - x} \right)dx} \]
Now adding like term:
\[ \Rightarrow I = \int_{0.5}^{1.5} {\left( {2 - x} \right)f\left( {2 - x} \right)dx} \]
Rewrite the integration:
\[ \Rightarrow I = \int_{0.5}^{1.5} {2f\left( {2 - x} \right)dx} - \int_{0.5}^{1.5} {xf\left( {2 - x} \right)dx} \]
Given that \[f\left( x \right) = f\left( {2 - x} \right)\].
\[ \Rightarrow I = \int_{0.5}^{1.5} {2f\left( x \right)dx} - \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
\[ \Rightarrow I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} - \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
We know that \[I = \int_{0.5}^{1.5} {xf\left( x \right)dx} \].
\[ \Rightarrow I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} - \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
\[ \Rightarrow I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} - I\]
Add I on both sides:
\[ \Rightarrow I + I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} \]
\[ \Rightarrow 2I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} \]
Divide both sides by 2:
\[ \Rightarrow \dfrac{{2I}}{2} = \dfrac{2}{2}\int_{0.5}^{1.5} {f\left( x \right)dx} \]
\[ \Rightarrow I = \int_{0.5}^{1.5} {f\left( x \right)dx} \]
Option ‘B’ is correct
Note: Students often make mistakes to solve the given definite integral. They used wrong property that is \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a - x} \right)dx} \]. But it is the wrong formula. The correct formula is \[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]. In the given integration the lower limit is not equal to zero. The correct property is \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \].
Formula Used:Definite integral property:
\[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx
Complete step by step solution:Given definite integral is
\[I = \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
Now applying the property of definite integral \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \]:
\[ \Rightarrow I = \int_{0.5}^{1.5} {\left( {1.5 + 0.5 - x} \right)f\left( {1.5 + 0.5 - x} \right)dx} \]
Now adding like term:
\[ \Rightarrow I = \int_{0.5}^{1.5} {\left( {2 - x} \right)f\left( {2 - x} \right)dx} \]
Rewrite the integration:
\[ \Rightarrow I = \int_{0.5}^{1.5} {2f\left( {2 - x} \right)dx} - \int_{0.5}^{1.5} {xf\left( {2 - x} \right)dx} \]
Given that \[f\left( x \right) = f\left( {2 - x} \right)\].
\[ \Rightarrow I = \int_{0.5}^{1.5} {2f\left( x \right)dx} - \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
\[ \Rightarrow I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} - \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
We know that \[I = \int_{0.5}^{1.5} {xf\left( x \right)dx} \].
\[ \Rightarrow I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} - \int_{0.5}^{1.5} {xf\left( x \right)dx} \]
\[ \Rightarrow I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} - I\]
Add I on both sides:
\[ \Rightarrow I + I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} \]
\[ \Rightarrow 2I = 2\int_{0.5}^{1.5} {f\left( x \right)dx} \]
Divide both sides by 2:
\[ \Rightarrow \dfrac{{2I}}{2} = \dfrac{2}{2}\int_{0.5}^{1.5} {f\left( x \right)dx} \]
\[ \Rightarrow I = \int_{0.5}^{1.5} {f\left( x \right)dx} \]
Option ‘B’ is correct
Note: Students often make mistakes to solve the given definite integral. They used wrong property that is \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a - x} \right)dx} \]. But it is the wrong formula. The correct formula is \[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]. In the given integration the lower limit is not equal to zero. The correct property is \[\int_b^a {f\left( x \right)dx} = \int_b^a {f\left( {a + b - x} \right)dx} \].
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