Problem

Which of the integrals $ \displaystyle \int^2_1 \…

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Problem 66 Hard Difficulty

Use properties of integrals, together with Exercises 27 and 28, to prove the inequality.

$ \displaystyle \int^{\pi/2}_0 x \sin x \,dx \le \frac{\pi^2}{8} $

Answer

$\int_{0}^{\pi / 2} x \sin x d x \leq \frac{\pi^{2}}{8}$

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Video Transcript

So they want us to prove this inequality here, Uh, using what we've done in 27 28. Um, so I'm just gonna write those results down, and we'll just kind of use them. So one thing that we might know, So let's do X sign of X here. Well, the reason why I would think they would want us to look at Exercises 27 28 is because we know that sign of X will always be less than or equal to one. So this here will always be less than or equal to just x times one. And now, since we have this inequality here, if we were to integrate each side, the left side should still be less than the right side. So this is going to be integral of zero two pi, half of x, Sign of X dx lesson opportunity. Integral from zero two pi, half of x dx. And now from number 27 this says, Well, it's just gonna be be square in my sights were all over to Let's go out to do that. So it will be pi squared over four minus zero all over to which is going to be pi squared over eight. So we have shown that this is less than or equal to pi squared or eight. So, um, so you passed your proof saying You love proof box and spotted face because you're glad you're done with it.

Bobby B.

University of North Texas

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Chapter 5

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Section 2

The Definite Integral

Problem

Which of the integrals $ \displaystyle \int^2_1 \…

Problem 66 Hard Difficulty

Use properties of integrals, together with Exercises 27 and 28, to prove the inequality.

$ \displaystyle \int^{\pi/2}_0 x \sin x \,dx \le \frac{\pi^2}{8} $

Answer

$\int_{0}^{\pi / 2} x \sin x d x \leq \frac{\pi^{2}}{8}$

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Calculus: Early Transcendentals

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Calculus: Early Transcendentals

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