prove that integral 0 to pi/2 sin x by x dx is greater than 1 and less than pi by 2

To prove that the integral \( \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \) is greater than 1 and less than \( \frac{\pi}{2} \), we can proceed as follows:

### Step 1: Establishing an Upper Bound

First, we observe that \( \frac{\sin x}{x} \leq 1 \) for all \( x \) in the interval \( [0, \frac{\pi}{2}] \). This is because:

\[
\frac{\sin x}{x} \leq 1 \quad \text{since} \quad \sin x \leq x \text{ for small } x, \text{ and for all } x \in [0, \frac{\pi}{2}].
\]

So, the integral \( \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \) can be bounded above by the integral of 1 over the same interval:

\[
\int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \leq \int_0^{\frac{\pi}{2}} 1 \, dx = \frac{\pi}{2}.
\]

This establishes the upper bound:
\[
\int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \leq \frac{\pi}{2}.
\]

### Step 2: Establishing a Lower Bound

To establish the lower bound, we can consider a comparison with the function \( \sin x \). Notice that \( \frac{\sin x}{x} \) is greater than or equal to \( \frac{2}{\pi} \) for \( x \) in the interval \( [0, \frac{\pi}{2}] \).

This is because the sine function is concave in the interval \( [0, \frac{\pi}{2}] \), and by Jensen's inequality, we have:

\[
\frac{\sin x}{x} \geq \frac{2}{\pi}
\]

Hence,

\[
\int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \geq \int_0^{\frac{\pi}{2}} \frac{2}{\pi} \, dx = 1.
\]

This establishes the lower bound:
\[
\int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \geq 1.
\]

### Conclusion

Combining the two bounds, we have:

\[
1 \leq \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \leq \frac{\pi}{2}.
\]

This proves that \( \int_0^{\frac{\pi}{2}} \frac{\sin x}{x} \, dx \) is greater than 1 and less than \( \frac{\pi}{2} \).

Would you like more details on any specific part of this proof, or do you have any questions?

Here are 5 related questions to explore:
1. How does Jensen's inequality apply to other integrals involving trigonometric functions?
2. Can we use similar techniques to bound integrals involving \(\cos x\) instead of \(\sin x\)?
3. How would the proof change if the integral were from 0 to \(\pi\) instead of 0 to \(\frac{\pi}{2}\)?
4. What are other methods to estimate the value of this integral more precisely?
5. How does the value of this integral compare to integrals of similar forms like \( \int_0^{\frac{\pi}{2}} \frac{\cos x}{x} \, dx \)?

**Tip:** Understanding inequalities like \( \sin x \leq x \) and \( \sin x \geq \frac{2x}{\pi} \) can be very useful for bounding integrals and sums in mathematical analysis.

Learn how to prove that the integral of sin(x)/x from 0 to pi/2 is greater than 1 and less than pi/2. This problem involves using Jensen's Inequality and understanding trigonometric inequalities. Suitable for advanced high school and undergraduate students.

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