# How do you use the properties of integrals to verify the inequality #intsinx/x# from pi/4 to pi/2 is less than or equal to #sqrt(2)/2#?

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To verify the inequality ( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin(x)}{x} , dx \leq \frac{\sqrt{2}}{2} ), we can compare it with a known integral and utilize the properties of integrals.

- Define ( f(x) = \frac{\sin(x)}{x} ).
- Note that ( f(x) ) is continuous on the interval ( [\frac{\pi}{4}, \frac{\pi}{2}] ) except possibly at ( x = 0 ).
- Since ( \sin(x) ) is non-negative on ( [\frac{\pi}{4}, \frac{\pi}{2}] ), we have ( 0 \leq f(x) ).
- Observe that ( \frac{\sqrt{2}}{2} ) is the maximum value of ( f(x) ) on ( [\frac{\pi}{4}, \frac{\pi}{2}] ) since ( \frac{\sin(x)}{x} ) is decreasing on this interval.
- Therefore, ( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin(x)}{x} , dx \leq \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sqrt{2}}{2} , dx = \frac{\sqrt{2}}{2} \cdot \frac{\pi}{2} = \frac{\pi\sqrt{2}}{4} ).

Thus, we have verified that ( \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sin(x)}{x} , dx \leq \frac{\sqrt{2}}{2} ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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