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Problem 71 Hard Difficulty

The rational number $ \frac{22}{7} $ has been used as an approximation to the number $ \pi $ since the time of Archimedes. Show that
$$ \int_0^1 \frac{x^4 (1 - x)^4}{1 + x^2}\ dx = \frac{22}{7} - \pi $$

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

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Video Transcript

Let's go ahead and evaluate the integral, starting with the left hand side. Let's just go ahead and distribute this. So let's expand this numerator and then denominators stays the same. So all we did there was expand. Now let's go ahead and simplify this. So this we should go ahead and do long division here. So let's go to the side. Do that. We've done this before. Can't a synthetic Yuri will need long division because of the quadratic. But after you do your long division over, So I'm making no here long division or polynomial division, I should say five x of the forth. So this is the quotient, and then you get a remainder of minus four and then the original denominator. So this becomes the general. As you see, we can just use upon rule whole bunch of times and then the very last one here, you write. Remember this. You hear? This will just be fourteen in verse. But if you feel about that fact, you can go ahead and just do it. Trips up here, That's equals tan data. So when we evaluate this exit, the seven over seven for exit the six over six find its five over five for X cubed over three for X for our plan. And then we have our entwine zero and one. So score the next patient plug those in, and then our ten of one is pi over four. So that's from plugging in the one. And then when we plug in zero, all the terms are zero. So go ahead and cancel those force and you get a minus pi up here and had combined the remaining fractions. You get twenty two over seven, and that's exactly what we wanted to prove. So that's we've evaluated the integral, and that's your final answer.

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Calculus: Early Transcendentals

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Problem

(a) Use integration by parts to show that, for an…

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Calculus: Early Transcendentals

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Watch More Solved Questions in Chapter 7

Video Transcript

James Stewart

Calculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Samuel Hannah

James StewartCalculus: Early Transcendentals

Grace He

Catherine Ross

Heather Zimmers

Samuel Hannah

James Stewart

Calculus: Early Transcendentals