Solution 03 04 2016 at 03 00am EST 1 pt Use the Midpoint Rule to approximate Z 5 1 pt Evaluate the denite integral Z 3 t ln t dt x3 dx 1 1 1 pt
Solution at am EST pt Use the Midpoint Rule to approximate Z pt Evaluate the denite integral Z t ln
Solution at am EST pt Use the Midpoint Rule to approximate Z pt Evaluate the
Use the Midpoint Rule to approximate Z pt Evaluate the denite integral Z t ln t dt x dx pt
Solution at am EST pt Use the Midpoint Rule to approximate Z
pt Evaluate the denite integral Z t ln t dt x dx pt
Solution at am EST pt Use the Midpoint Rule to
Solution at am EST pt
(Solution) 03/04/2016 at 03:00am EST. (1 pt) Use the Midpoint Rule to approximate Z 5. (1 pt) Evaluate the denite integral. Z 3 t ln(t) dt = x3 dx 1 1. (1 pt)...

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Hi, I need to solve the first 15 questions. ThanksMohammed Yaaseen Gomdah E WeBWorK assignment due : 03/04/2016 at 03:00am EST. 1. (1 pt) Use the Midpoint Rule to approximate Z 5 . 5 - 1 . 5 x 3 dx with n = 7. 2. (1 pt) Consider the function f ( x ) = - x 2 2 - 5. In this problem you will calculate Z 4 0 ± - x 2 2 - 5 ² dx by us- ing the de?nition Z b a f ( x ) dx = lim n ? ? " n ? i = 1 f ( x i ) ? x # The summation inside the brackets is R n which is the Rie- mann sum where the sample points are chosen to be the right- hand endpoints of each sub-interval. Calculate R n for f ( x ) = - x 2 2 - 5 on the interval [ 0 , 4 ] and write your answer as a function of n without any summation signs. You will need the summation formulas on page 383 of your textbook (page 364 in older texts). R n = lim n ? ? R n = 3. (1 pt) Let Z - 4 - 7 f ( x ) dx = 1 , Z - 6 - 7 f ( x ) dx = 9 , Z - 4 - 5 f ( x ) dx = 1. Find Z - 5 - 6 f ( x ) dx = and Z - 6 - 5 ( 1 f ( x ) - 9 ) dx = 4. (1 pt) If f ( x ) = Z x 2 3 t 4 dt then f 0 ( x ) = f 0 ( 5 ) = 5. (1 pt) Given f ( x ) = Z x 0 t 2 - 4 1 + cos 2 ( t ) dt At what value of x does the local max of f ( x ) occur? x = 6. (1 pt) Evaluate the de?nite integral. Z 1 0 x 2 5 ? e x dx = 7. (1 pt) Evaluate the de?nite integral. Z 3 1 ? t ln ( t ) dt = 8. (1 pt) Use the Fundamental Theorem of Calculus to carry out the following differentiation: d d x Z ? x 1 t t d t = . 9. (1 pt) Integration by Parts: This is the most important in- tegration technique we’ve discussed in this class. It has a wide range of applications beyond increasing our list of integration rules. R z 3 ln z d z = . R e t cos t d t = . R 2 ? 0 sin ( x ) sin ( x + 1 ) d x = . 10. (1 pt) Use integration by parts to evaluate the integral. Z xe 4 x dx Answer: + C 11. (1 pt) Use integration by parts to evaluate the de?nite integral. Z 7 1 ? t ln tdt Answer: 12. (1 pt) Evaluate the de?nite integral. Z 5 3 ln x 42 dx Answer: 13. (1 pt) Evaluate the inde?nite integral. Z ln ( x 2 + 9 x + 18 ) dx 1

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