Problem 108 - Calculus Volume 1

Question

In the following exercises, use the comparison theorem. Show that \(\int_{0}^{\pi / 2} \sin t d t \geq \frac{\pi}{4} .\) (Hint: sin \(t \geq \frac{2 t}{\pi}\) over \(\left[0, \frac{\pi}{2}\right])\)

Step-by-Step Solution

Answer

The inequality \(\int_{0}^{\pi/2} \sin t\, dt \geq \frac{\pi}{4}\) holds true as both integrals are evaluated and compared.

Step 1: Understanding the Comparison Theorem

The Comparison Theorem is useful for determining the convergence or inequality of improper integrals by comparing them with simpler integrals. If a function \(f(t)\) is greater than or equal to \(g(t)\) on an interval, then \(\int f(t) \, dt \geq \int g(t) \, dt\) over that interval, provided both integrals are finite.

Step 2: Use the Given Inequality

We are given the inequality \(\sin t \geq \frac{2t}{\pi}\) for \(t\) in the interval \([0, \frac{\pi}{2}]\). This means that for all \(t\) in this interval, the value of \(\sin t\) is greater than or equal to \(\frac{2t}{\pi}\).

Step 3: Set Up the Inequality

Using the comparison theorem, we know if \(\sin t \geq \frac{2t}{\pi}\), then \[\int_{0}^{\pi/2} \sin t \, dt \geq \int_{0}^{\pi/2} \frac{2t}{\pi} \, dt.\] This sets up our inequality. Our goal is to evaluate both integrals to show the inequality holds.

Step 4: Evaluate \(\int_{0}^{\pi/2} \sin t \, dt\)

To evaluate the integral of \(\sin t\), \[\int_{0}^{\pi/2} \sin t \, dt = \left[ -\cos t \right]_{0}^{\pi/2} = -\cos \left(\frac{\pi}{2}\right) + \cos(0) = 0 + 1 = 1.\]

Step 5: Evaluate \(\int_{0}^{\pi/2} \frac{2t}{\pi} \, dt\)

To evaluate the integral of \(\frac{2t}{\pi}\), we perform the following: \[\int_{0}^{\pi/2} \frac{2t}{\pi} \, dt = \frac{2}{\pi} \cdot \left[ \frac{t^2}{2} \right]_{0}^{\pi/2} = \frac{2}{\pi} \cdot \left( \frac{(\pi/2)^2}{2} - 0 \right) = \frac{2}{\pi} \cdot \frac{\pi^2}{8} = \frac{\pi}{4}.\]

Step 6: Conclusion

Since \(\int_{0}^{\pi/2} \sin t \, dt = 1\) and \(\int_{0}^{\pi/2} \frac{2t}{\pi} \, dt = \frac{\pi}{4}\), we observe that 1 is indeed greater than \(\frac{\pi}{4}\). Therefore, the inequality \(\int_{0}^{\pi / 2} \sin t \, dt \geq \frac{\pi}{4}\) holds true.

Key Concepts

sin t inequality

Understanding the inequality of functions is crucial in calculus. In this exercise, we deal with an inequality involving the sine function:

• Given: \( \sin t \geq \frac{2t}{\pi} \) over the interval \( [0, \frac{\pi}{2}] \).
• This tells us that values of \( \sin t \) are always greater than or equal to those of \( \frac{2t}{\pi} \) within the specified range.

The intuition behind this inequality is simpler than it seems. As \( t \) moves from 0 to \( \frac{\pi}{2} \), the sine function climbs from \( \sin 0 = 0 \) to \( \sin \frac{\pi}{2} = 1 \). Meanwhile, the linear function \( \frac{2t}{\pi} \) rises from 0 to just \( 1 \). By examining points and the general behavior of these functions within this interval, we see how and why the inequality makes sense.

integral evaluation

Evaluating the definite integral is key to solving this problem. We need to find \( \int_{0}^{\pi/2} \sin t \, dt \). Here's how it's done:

• First, find the antiderivative of \( \sin t \), which is \( -\cos t \).
• Then, apply the limits from 0 to \( \frac{\pi}{2} \).
• So, \( \int_{0}^{\pi/2} \sin t \, dt = \left[ -\cos t \right]_{0}^{\pi/2} = 0 + 1 = 1 \).

Integrals give us the area under the curve of a function over an interval. For \( \sin t \), the area from 0 to \( \frac{\pi}{2} \) equals 1, showing how much the function accumulates over this specific range. This technique is widely used across different kinds of problems in calculus.

improper integrals

Improper integrals occur when the limits of integration extend to infinity or when integrands become undefined within the limits. However, in this exercise, we deal with a definite integral that is not inherently improper.

• Our integral has finite intervals: \([0, \frac{\pi}{2}]\).
• No discontinuities or infinite discontinuities are involved in this range.

Nonetheless, the concept of improper integrals is essential to understand as it helps grasp how and when to apply certain theorems and comparison techniques, such as the Comparison Theorem used in this exercise.

convergence of integrals

The convergence of an integral refers to whether it has a finite value. In this case, both integrals are convergent:

• \( \int_{0}^{\pi/2} \sin t \, dt = 1 \), a finite value.
• \( \int_{0}^{\pi/2} \frac{2t}{\pi} \, dt = \frac{\pi}{4} \), also finite.

The Comparison Theorem assists in establishing this convergence. Given the inequality \( \sin t \geq \frac{2t}{\pi} \), we are assured of the convergence at least to \( \frac{\pi}{4} \). By having both integrals converge, mathematical relationships regarding their differences or ratios can be clearly defined and manipulated, verifying inequalities or approximations as needed in calculus.