# P 1 p 2 p 3 p n n 3 2 click next to see the answer

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p 1 + p 2 + p 3 + ° ° ° + p n n 3 / 2 . Click Next to see the answer. Ziad Z. Adwan Lecture 16 & Lab 16 32 / 57
Using the Fundamental Theorem of Calculus to °nd Limits Example ( Using the Fundamental Theorem of Calculus to Compute Limits) Use FTC1 & Riemann Sums to evaluate: lim n ! p 1 + p 2 + p 3 + ° ° ° + p n n 3 / 2 . Click Next to see the answer. p 1 + p 2 + °°° + p n n 3 / 2 = ºq 1 n + q 2 n + ° ° ° + p n n » ´ 1 n µ = n j = 1 q j n ´ 1 n µ = n j = 1 f ( c j ) 4 x j , with f ( x ) = p x on [ 0 , 1 ] , c j = j / n & the partition 4 is a regular partition with width 4 x j = ( j / n ) ± (( j ± 1 ) / n ) = 1 / n . Ziad Z. Adwan Lecture 16 & Lab 16 32 / 57
Using the Fundamental Theorem of Calculus to °nd Limits Example ( Using the Fundamental Theorem of Calculus to Compute Limits) Use FTC1 & Riemann Sums to evaluate: lim n ! p 1 + p 2 + p 3 + ° ° ° + p n n 3 / 2 . Click Next to see the answer. p 1 + p 2 + °°° + p n n 3 / 2 = ºq 1 n + q 2 n + ° ° ° + p n n » ´ 1 n µ = n j = 1 q j n ´ 1 n µ = n j = 1 f ( c j ) 4 x j , with f ( x ) = p x on [ 0 , 1 ] , c j = j / n & the partition 4 is a regular partition with width 4 x j = ( j / n ) ± (( j ± 1 ) / n ) = 1 / n . Thus, by De°nition of De°nite Integral as the Limit of a Riemann Sum: lim n ! p 1 + p 2 + p 3 + ° ° ° + p n n 3 / 2 = Z 1 0 p xdx FTC1 = ² 2 3 x 3 / 2 ³ 1 0 = 2 3 . ° Ziad Z. Adwan Lecture 16 & Lab 16 32 / 57
Using the Fundamental Theorem of Calculus to °nd Limits Example ( Using the Fundamental Theorem of Calculus to Compute Limits) Use FTC1 & Riemann Sums to evaluate: lim n ! 1 5 + 2 5 + 3 5 + ° ° ° + n 5 n 6 . Click Next to see the answer. Ziad Z. Adwan Lecture 16 & Lab 16 33 / 57
Using the Fundamental Theorem of Calculus to °nd Limits Example ( Using the Fundamental Theorem of Calculus to Compute Limits) Use FTC1 & Riemann Sums to evaluate: lim n ! 1 5 + 2 5 + 3 5 + ° ° ° + n 5 n 6 . Click Next to see the answer. 1 5 + 2 5 + °°° + n 5 n 6 = º ´ 1 n µ 5 + ´ 2 n µ 5 + ° ° ° + ´ n n µ 5 » ´ 1 n µ = n j = 1 º j n » 5 ´ 1 n µ = n j = 1 f ( c j ) 4 x j , with f ( x ) = x 5 on [ 0 , 1 ] , c j = j / n & the partition 4 is a regular partition with width 4 x j = ( j / n ) ± (( j ± 1 ) / n ) = 1 / n . Ziad Z. Adwan Lecture 16 & Lab 16 33 / 57
Using the Fundamental Theorem of Calculus to °nd Limits Example ( Using the Fundamental Theorem of Calculus to Compute Limits) Use FTC1 & Riemann Sums to evaluate: lim n ! 1 5 + 2 5 + 3 5 + ° ° ° + n 5 n 6 . Click Next to see the answer. 1 5 + 2 5 + °°° + n 5 n 6 = º ´ 1 n µ 5 + ´ 2 n µ 5 + ° ° ° + ´ n n µ 5 » ´ 1 n µ = n j = 1 º j n » 5 ´ 1 n µ = n j = 1 f ( c j ) 4 x j , with f ( x ) = x 5 on [ 0 , 1 ] , c j = j / n & the partition 4 is a regular partition with width 4 x j = ( j / n ) ± (( j ± 1 ) / n ) = 1 / n . Thus, by De°nition of De°nite Integral as the Limit of a Riemann Sum: lim n ! 1 5 + 2 5 + 3 5 + ° ° ° + n 5 n 6 = Z 1 0 x 5 dx FTC1 = ² 1 6 x 6 ³ 1 0 = 1 6 . ° Ziad Z. Adwan Lecture 16 & Lab 16 33 / 57
Average Value of a function on an Interval De°nition ( Average Value of a function on an Interval) If f is integrable on [ a , b ] , then the average value of f on [ a , b ] is de°ned by: Ave ( f ) + 1 b ± a Z b a f ( x ) dx . Note (in the °gure below) that the area of the region under the graph of f is equal to the area of the rectangle whose height is Ave ( f ) . Can you explain why?
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