Use the properties of integrals to verify the inequality without evaluating the integrals
I am writing the inequality below.

Question

Use the properties of integrals to verify the inequality without evaluating the integrals
I am writing the inequality below.

anonymous · Answer

$$\int\limits_{0}^{1}\sqrt{1+x^2} \le \int\limits_{0}^{1} \sqrt{1+x}$$

anonymous · Answer

if you let f(x) be the function inside the first integral and g(x) be the function inside the second integral then I know the next step is proving that
f(x) < g(x)
but for some reason my brain is just freezing here and I just cannot see how to prove that statement

anonymous · Answer

sorry the less than symbol in my last post is suppose to be a less than or equal to symbol

sirm3d · Answer

$$x^2<x \text{ for } 0<x<1$$

anonymous · Answer

I know it is true but the inequality I get when I try  to prove it is just x < 1, wait a second g(x) does not exist past -1 that is where you get the second half of the inequality from sorry I should have seen that sooner.