# Use the properties of integrals to verify the inequality without evaluating the integrals. integrate from `int_0^1 sqrt(1+x^2)dx lt int_0^1 sqrt(1+x)dx`

You should remember that `a<b`  and `f(x) < g(x)` , then `int_a^b f(x) dx < int_a^b g(x) dx.`

Notice that 0<1 and you need to verify if `sqrt(1+x^2) < sqrt(1+x) ` over the interval [0,1] such that:

If `x in [0,1] => x^2 < x => x^2 + 1 <...

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You should remember that `a<b`  and `f(x) < g(x)` , then `int_a^b f(x) dx < int_a^b g(x) dx.`

Notice that 0<1 and you need to verify if `sqrt(1+x^2) < sqrt(1+x) ` over the interval [0,1] such that:

If `x in [0,1] => x^2 < x => x^2 + 1 < x + 1 => sqrt(x^2 + 1)<sqrt(x+1)`

Since `sqrt(x^2 + 1)<sqrt(x+1),`  then you may use the property of integrals stated above such that:

`sqrt(x^2 + 1)<sqrt(x+1) => int_0^1 sqrt(x^2 + 1) dx < int_0^1 sqrt(x+1)dx`

Hence, using the properties of integrals yields that inequality `int_0^1 sqrt(x^2 + 1) dx < int_0^1 sqrt(x+1) dx`  holds.

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