`int_0^1sqrt(1 + x^2)dx <= int_0^1sqrt(1 + x)dx` Use the properties of integrals to verify the inequality without evaluating the integrals.

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Chapter 5, 5.2 - Problem 56 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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You need to check if` int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx` , using mean value theorem, such that:

`int_a^b f(x)dx = (b-a)f(c), ` where `c in (a,b)`

`int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx => int_0^1 sqrt(1+x^2)dx - int_0^1sqrt(1+x)dx <= 0`

`int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx <= 0`

According to mean value theorem yields:

`int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx = (1-0)f(c), c in (0,1)`

You need to verify the monotony of the function `f(x) = sqrt(1+x^2)dx - sqrt(1+x), ` such that:

`f'(x) = x/(sqrt(1+x^2)) - 1/(2sqrt(1+x))`

`f'(x) = (2x*sqrt(1+x)-sqrt(1+x^2)) /(2sqrt(1+x^2)(1+x))`

`f'(x) = 0 => 2x*sqrt(1+x)-sqrt(1+x^2) = 0 => 2x*sqrt(1+x)=sqrt(1+x^2)`

Raising to square:

 

`4x^2(1+x) = 1+x^2`

`4x^2 + 4x^3 - 1 - x^2 = 0`

`4x^3 + 3x^2 - 1 >0 for x in (0,1) =>`   f(x) increases on (0,1)

For `c in (0,1) => 0<c<1 => f(0)<f(c)<f(1)`

`f(0) = sqrt1 - sqrt1 = 0`

`f(1) = sqrt(1+1) - sqrt(1+1)=0`

Since` f(c) < f(1) = 0 => f(c) < 0` . Since `f(c) = int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx`  , then` int_0^1 (sqrt(1+x^2)dx - sqrt(1+x))dx< 0.`

Hence, checking if the inequality holds, using mean value theorem, yields `int_0^1 sqrt(1+x^2)dx <= int_0^1sqrt(1+x)dx` is verified.

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