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Question: The value of \(\int _ { - 1 } ^ { 1 } \left( \sqrt { 1 + x + x ^ { 2 } } - \sqrt { 1 - x + x ^ { 2 }...

The value of ∫−11(1+x+x2−1−x+x2)dx\int _ { - 1 } ^ { 1 } \left( \sqrt { 1 + x + x ^ { 2 } } - \sqrt { 1 - x + x ^ { 2 } } \right) d x is

A

0

B

1

C

−1- 1

D

None of these

Answer

0

Explanation

Solution

Let f(x)=1+x+x2−1−x+x2f ( x ) = \sqrt { 1 + x + x ^ { 2 } } - \sqrt { 1 - x + x ^ { 2 } }.

Then f(−x)=1−x+x2−1+x+x2=−f(x)f ( - x ) = \sqrt { 1 - x + x ^ { 2 } } - \sqrt { 1 + x + x ^ { 2 } } = - f ( x )

Hence f(x)f ( x ) is an odd function and so

∫−11f(x)dx=0\int _ { - 1 } ^ { 1 } f ( x ) d x = 0.