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There's a bonus question on my homework that asks why the integral from

a to b sqrt(1+1/x^4)dx

is greater than the integral from

a to b (1dx).

I know they come close to each other but I can't seem to find a way to explain it in words. Any ideas?

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    $\sqrt{x}$ is increasing, maybe that's enough. To conclude that because $\frac{1}{x^4}>0$ we have $1+\frac{1}{x^4}>1$ and finally $\sqrt{1+\frac{1}{x^4}}>\sqrt{1}$. – 2017-02-28

2 Answers 2

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Notice that $1 + 1/x^4$ is always greater than $1$ because $1/x^4$ is positive. Since that's greater than 1 we know $\sqrt{1 + 1/x^4} > 1$ so $\int_1^b \sqrt{1 + 1/x^4} dx > \int_a^b 1 dx$.

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Since $$\sqrt{1+\frac{1}{x^4}}-1\gt0\quad\forall x\in\mathbb{R},$$ $$\int_{a}^{b}\left(\sqrt{1+\frac{1}{x^4}}-1\right)dx\gt0$$ By linearity of the integral, $$\int_{a}^{b}\sqrt{1+\frac{1}{x^4}}dx\gt\int_{a}^{b}dx.$$