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You need to use the mean value theorem such that:
`int_a^b f(x)dx = (b-a)f(c), c in (a,b)`
`int_(-1)^1 sqrt(1+x^2)dx = (1+1)f(c) = 2f(c)`
You need to verify the monotony of the function `f(x) = sqrt(1+x^2),` such that:
`f'(x) = x/(sqrt(1+x^2))`
Since the function is even, then `int_(-1)^1 sqrt(1+x^2)dx = 2int_0^1 sqrt(1+x^2)dx` . Notice that f(x) increases on (0,1).
Hence, if 0<c<1, then `f(0)<f(c)<f(1).`
Evaluate f`(0) = sqrt 1 ` and `f(1) = sqrt2` , such that:
`sqrt1 <f(c) < sqrt 2`
Multiply by 2:
Replace 2f(c) by` int_(-1)^1 sqrt(1+x^2)dx` such that:
`2<int_(-1)^1 sqrt(1+x^2)dx <2sqrt2`
Hence, the inequality is verified using mean value theorem, without evaluating the integral.
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