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Decide if integral is positive or negative without calculating the integral.

$\int\limits^{1}_{3} (x^2+2)\,\mathrm{d}x$

Here my thought process is:

Since lower bound has a higher value than higher bound, I have to switch and will get -$\int\limits^{3}_{1} (x^2+2)\,\mathrm{d}x$ and since $(x^2+2)$ is always positive, so it will be negative? Other solutions? What if the function isn't always positive, any other properties I can use here?

Decide which integral is bigger of the two without calculating the integral.

$\int\limits^{1}_{0} \sqrt{x}\,\mathrm{d}x$ and $\int\limits^{1}_{0} x^2\,\mathrm{d}x$

My thoughts:

There is a property if f(x) <= g(x) then integrals with those functions with same bounds also act the same way.

But if the first one is integral of f(x) here then it doesn't seem to apply, because $\sqrt{x}$ <= $x^2$ and if I do calculate the integrals I get that the $\int\limits^{1}_{0} x^2\,\mathrm{d}x$ is bigger. I am confused. Help?

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    The first one is negative, as you described. – 2017-01-17
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    Your thinking is right, except for that $\sqrt{x}$ <= $x^2$ bit (remember that $x \in [0,1]$). – 2017-01-17
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    Picking one example $x \in [0, 1]$, how about $1/4$. Which is bigger: $\sqrt{1/4}$ or $(1/4)^2$? – 2017-01-17
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    Oh yeah, good point. Thank you. – 2017-01-17

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