
What is the value of \[\int_{ - 1}^1 {\left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)dx} \]?
A. 0
B. 1
C. -1
D. None of these
Answer
164.1k+ views
Hint: The given differential equation is known as a definite integral. First, we check whether the given function is an odd function or an even function. Then we will apply an odd function on the definite integral formula to calculate the value of the given integral.
Formula Used:Property of the definite integral:
\[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\end{array}} \right.\]
Complete step by step solution:Given definite integral is
\[\int_{ - 1}^1 {\left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)dx} \]
Assume that, \[f\left( x \right) = \sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} \].
To check whether \[f\left( x \right)\] is an odd function or even function, we will put x = -x in \[f\left( x \right)\].
\[f\left( { - x} \right) = \sqrt {1 + \left( { - x} \right) + {{\left( { - x} \right)}^2}} - \sqrt {1 - \left( { - x} \right) + {{\left( { - x} \right)}^2}} \]
\[ \Rightarrow f\left( { - x} \right) = \sqrt {1 - x + {x^2}} - \sqrt {1 + x + {x^2}} \]
Taking common negative sign:
\[ \Rightarrow f\left( { - x} \right) = - \left[ {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right]\]
Now putting \[\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} = f\left( x \right)\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
The given function is an odd function.
Now applying the formula \[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = f\left( x \right)}\end{array}} \right.\] in the given definite integral:
\[\int_{ - 1}^1 {\left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)dx} = 0\]
Option ‘A’ is correct
Note: Students often make mistakes to apply the formula. They use \[\int_{ - a}^a {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)\] which is incorrect formula. The correct formula is \[\int\limits_{ - a}^a {f\left( x \right)dx = 0\,} {\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)\].
Formula Used:Property of the definite integral:
\[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\end{array}} \right.\]
Complete step by step solution:Given definite integral is
\[\int_{ - 1}^1 {\left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)dx} \]
Assume that, \[f\left( x \right) = \sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} \].
To check whether \[f\left( x \right)\] is an odd function or even function, we will put x = -x in \[f\left( x \right)\].
\[f\left( { - x} \right) = \sqrt {1 + \left( { - x} \right) + {{\left( { - x} \right)}^2}} - \sqrt {1 - \left( { - x} \right) + {{\left( { - x} \right)}^2}} \]
\[ \Rightarrow f\left( { - x} \right) = \sqrt {1 - x + {x^2}} - \sqrt {1 + x + {x^2}} \]
Taking common negative sign:
\[ \Rightarrow f\left( { - x} \right) = - \left[ {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right]\]
Now putting \[\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} = f\left( x \right)\]
\[ \Rightarrow f\left( { - x} \right) = - f\left( x \right)\]
The given function is an odd function.
Now applying the formula \[\int_{ - a}^a {f\left( x \right)dx} = \left\{ {\begin{array}{*{20}{c}}{0\,{\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)}\\{2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = f\left( x \right)}\end{array}} \right.\] in the given definite integral:
\[\int_{ - 1}^1 {\left( {\sqrt {1 + x + {x^2}} - \sqrt {1 - x + {x^2}} } \right)dx} = 0\]
Option ‘A’ is correct
Note: Students often make mistakes to apply the formula. They use \[\int_{ - a}^a {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx\,} {\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)\] which is incorrect formula. The correct formula is \[\int\limits_{ - a}^a {f\left( x \right)dx = 0\,} {\rm{if}}\,f\left( { - x} \right) = - f\left( x \right)\].
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