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Pre-Calc Homework Solutions 213

# Pre-Calc Homework Solutions 213 - Section 5.3 213(d lim n k...

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(d) lim n k 5 1 11 } n k } 2 2 ? } 1 n } 2 5 lim n } n ( n 1 1 6 ) n ( 3 2 n 1 1) } 5 lim n 5 } 2 6 } 5 } 1 3 } (e) Since E 1 0 x 2 dx equals the limit of any Riemann sum over the interval [0, 1] as n approaches , part (d) proves that E 1 0 x 2 dx 5 } 1 3 } . Section 5.3 Definite Integrals and Antiderivatives (pp. 268–276) Exploration 1 How Long is the Average Chord of a Circle? 1. The chord is twice as long as the leg of the right triangle in the first quadrant, which has length ˇ r 2 w 2 w x w 2 w by the Pythagorean Theorem. 2. Average value 5 } r 2 1 ( 2 r ) } E r 2 r 2 ˇ r 2 w 2 w x w 2 w dx . 3. Average value 5 } 2 2 r } E r 2 r ˇ r 2 w 2 w x w 2 w dx 5 } 1 r } ? (area of semicircle of radius r ) 5 } 1 r } ? } p 2 r 2 } 5 } p 2 r } 4. Although we only computed the average length of chords perpendicular to a particular diameter, the same computation applies to any diameter. The average length of a chord of a circle of radius r is } p 2 r } . 5. The function y 5 2 ˇ r 2 w 2 w x w 2 w is continuous on [ 2 r , r ], so the Mean Value Theorem applies and there is a c in [ a , b ] so that y ( c ) is the average value } p 2 r } . Exploration 2 Finding the Derivative of an Integral Pictures will vary according to the value of x chosen. (Indeed, this is the point of the exploration.) We show a
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