
What is the value of \[\int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \]?
A. \[\dfrac{\pi }{4}\]
B. \[\dfrac{\pi }{2}\]
C. \[{e^{\dfrac{{{\pi ^2}}}{{16}}}}\]
D. \[{e^{\dfrac{{{\pi ^2}}}{4}}}\]
Answer
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Hint: To solve the given definite integral we will apply the property of the definite integral. As the lower limit of the definite integral is zero, thus we will apply \[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]. Then add the integral and simplify to get simplest form the integration and solve it.
Formula Used:Definite integral property:
\[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]
Formula of integration:
\[\int {dx} = x + c\]
Complete step by step solution:Given definite integral is
\[\int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \]
Assume that, \[I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \] …….(i)
Now applying the property of definite integral \[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]:
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - \dfrac{\pi }{2} + x} \right)}^2}}}}}dx} \]
Now subtracting like term:
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}} + {e^{{x^2}}}}}dx} \] ….. (ii)
Adding equation (i) and (ii)
\[I + I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} + \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}} + {e^{{x^2}}}}}dx} \]
Rewrite the integration:
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \]
Now cancel out like terms:
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {1dx} \]
Integrate the right hand side:
\[ \Rightarrow 2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]
Now putting lower limit and upper limit
\[ \Rightarrow 2I = \dfrac{\pi }{2} - 0\]
Divide both sides by 2:
\[ \Rightarrow I = \dfrac{\pi }{4}\]
Option ‘A’ is correct
Note: Students often make mistake when they solve a definite integral. They put integrating factor in the step \[2I = \left[ {x + c} \right]_0^{\dfrac{\pi }{2}}\]. But it is incorrect because we don’t need to put integrating constant when we solve definite integral. The correct way is \[2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\].
Formula Used:Definite integral property:
\[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]
Formula of integration:
\[\int {dx} = x + c\]
Complete step by step solution:Given definite integral is
\[\int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \]
Assume that, \[I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \] …….(i)
Now applying the property of definite integral \[\int_0^a {f\left( x \right)dx} = \int_0^a {f\left( {a - x} \right)dx} \]:
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - \dfrac{\pi }{2} + x} \right)}^2}}}}}dx} \]
Now subtracting like term:
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}} + {e^{{x^2}}}}}dx} \] ….. (ii)
Adding equation (i) and (ii)
\[I + I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} + \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}} + {e^{{x^2}}}}}dx} \]
Rewrite the integration:
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {\dfrac{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}{{{e^{{x^2}}} + {e^{{{\left( {\dfrac{\pi }{2} - x} \right)}^2}}}}}dx} \]
Now cancel out like terms:
\[ \Rightarrow 2I = \int_0^{\dfrac{\pi }{2}} {1dx} \]
Integrate the right hand side:
\[ \Rightarrow 2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\]
Now putting lower limit and upper limit
\[ \Rightarrow 2I = \dfrac{\pi }{2} - 0\]
Divide both sides by 2:
\[ \Rightarrow I = \dfrac{\pi }{4}\]
Option ‘A’ is correct
Note: Students often make mistake when they solve a definite integral. They put integrating factor in the step \[2I = \left[ {x + c} \right]_0^{\dfrac{\pi }{2}}\]. But it is incorrect because we don’t need to put integrating constant when we solve definite integral. The correct way is \[2I = \left[ x \right]_0^{\dfrac{\pi }{2}}\].
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