The rational number 22/7 has been used as an approximation to the number since the time...

 

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The rational number 22/7 has been used as an approximation to the number since the time of Archimedes. First, show that: L x(1-x)4 1+x dx = 22 7 - . We will now use properties of this integral to see how good of an approximation of the rational number 22/7 is. First, notice that the integrand is positive over the interval of integration (0, 1). Therefore, by properties of definite integrals (see your notes from MATH 144), it follows that L x(1-x)4 1+x dx > 0, that is, 22/7 >. But how close is 22/7 to the right answer? We use comparison tests for definite integrals. As the denominator 1 + x > 1 for all x = (0, 1), it follows that x(1-x)4 1+x < x4 (1-x)4 for all x = (0, 1). In other words, if we replace the denominator by 1 (i.e. omit the x2 term), then the new integrand becomes larger than the original one for all x = (0, 1). Therefore, by properties of definite integrals, x(1-x)4 1+x dx < x(1 - x) dx. Similarly, since 1 + x2 <2 for all x = (0, 1), if we replace the denominator by 2, then the new integrand becomes smaller than the original one for all x = (0, 1). Thus L x*(1-x)+ 1+x x(1-x)4 dx > dx. 2 Evaluate these two integrals, to conclude that: 1 22 1 1260 630 Manipulate this inequality to show that: 3.14127... < < < 3.14206... while the actual value of x is 3.14159.... Isn't that cool? Are you impressed with this bound on the value of ? The rational number 22/7 has been used as an approximation to the number since the time of Archimedes. First, show that: L x(1-x)4 1+x dx = 22 7 - . We will now use properties of this integral to see how good of an approximation of the rational number 22/7 is. First, notice that the integrand is positive over the interval of integration (0, 1). Therefore, by properties of definite integrals (see your notes from MATH 144), it follows that L x(1-x)4 1+x dx > 0, that is, 22/7 >. But how close is 22/7 to the right answer? We use comparison tests for definite integrals. As the denominator 1 + x > 1 for all x = (0, 1), it follows that x(1-x)4 1+x < x4 (1-x)4 for all x = (0, 1). In other words, if we replace the denominator by 1 (i.e. omit the x2 term), then the new integrand becomes larger than the original one for all x = (0, 1). Therefore, by properties of definite integrals, x(1-x)4 1+x dx < x(1 - x) dx. Similarly, since 1 + x2 <2 for all x = (0, 1), if we replace the denominator by 2, then the new integrand becomes smaller than the original one for all x = (0, 1). Thus L x*(1-x)+ 1+x x(1-x)4 dx > dx. 2 Evaluate these two integrals, to conclude that: 1 22 1 1260 630 Manipulate this inequality to show that: 3.14127... < < < 3.14206... while the actual value of x is 3.14159.... Isn't that cool? Are you impressed with this bound on the value of ?