f x e x 5 0 x 2 f x e x 10 f x e x 10 0 x 2 M 4 4 f x i \u0394 x f x 1 f x 2 f x 3 f

F x e x 5 0 x 2 f x e x 10 f x e x 10 0 x 2 m 4 4 f x

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f ( x ) = e x 5 , 0 x 2, f ( x ) = e x 10 f ( x ) = e x 10 , 0 x 2, M 4 = , 4 f ( x i ) Δ x = [ f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + f ( x 4 )] Δ x i = 1 x 1 , x 2 , x 3 , and x 4 [0, 2]. [0, 2] Δ x = (No Response) 1/2 .
11/17/14, 2:15 PM Math 125 HW_1C Page 2 of 15 2. 6/6 points | Previous Answers SCalcET7 5.2.004. (a) Find the Riemann sum for with six terms, taking the sample points to be right endpoints. (Round your answers to six decimal places.) R 6 = 2.776802 2.776802 (b) Repeat part (a) with midpoints as the sample points. M 6 = 5.130861 5.130861 Solution or Explanation Click to View Solution 3. 6/6 points | Previous Answers SCalcET7 5.2.007. A table of values of an increasing function f is shown. Use the table to find lower and upper estimates for lower estimate -76 -76 upper estimate 16 16 x 10 14 18 22 26 30 9 Solution or Explanation Click to View Solution f ( x ) = 5 sin x , 0 x 3 π /2, f ( x ) dx . 30 10 f ( x ) 14 9 1 1 4
11/17/14, 2:15 PM Math 125 HW_1C Page 3 of 15 4. 5/5 points | Previous Answers SCalcET7 5.2.009. Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. -3.0484 -3.0484 Solution or Explanation so the endpoints are 0, 14 , 28 , 42 , and 56 , and the midpoints are 7 , 21 , 35 , and 49 . The Midpoint Rule gives sin dx , n = 4 56 x 0 Δ x = ( 56 0)/4 = 14 , sin dx 56 x 4 f ( x i ) Δ x = 14 (sin + sin + sin + sin ) 14 ( 0.2177 ) = 3.0484 . i = 1 7 21 35 49 0
11/17/14, 2:15 PM Math 125 HW_1C Page 4 of 15 5. 5/5 points | Previous Answers SCalcET7 5.2.017.MI. Express the limit as a definite integral on the given interval. Master It Express the limit as a definite integral on the given interval. Part 1 of 2 On the interval [ a , b ], the limit gives us the integral For we have Solution or Explanation Click to View Solution 6. 9/9 points | Previous Answers SCalcET7 5.2.034.MI. The graph of g consists of two straight lines and a semicircle. Use it to evaluate each integral. lim n n x i ln( 1 + x i 2 ) Δ x , [ 2 , 6 ] i = 1 dx 6 2 lim n n x i ln( 1 + x i 4 ) Δ x , [ 1 , 4 ] i = 1 lim n n f ( x i ) Δ x i = 1 . f ( x ) dx b a Δ x , lim n n x i ln( 1 + x i 4 ) i = 1 f ( x ) = (No Response) .
11/17/14, 2:15 PM Math 125 HW_1C Page 5 of 15 (a) Master It The graph of g ( x ) consists of two straight lines and a semicircle. Use it to evaluate the integral. g ( x ) dx 10 0 g ( x ) dx 8 0
11/17/14, 2:15 PM Math 125 HW_1C Page 6 of 15 Step 1 of 2 If g ( x ) is positive, then the integral corresponds to the area beneath g ( x ) and above the x - axis over the interval [ a , b ]. On [0, 8 ], the function g ( x ) is above the x -axis and is therefore positive. Thus, equals the area of the triangle created by the function, the x -axis, and the y -axis. This triangle is a right triangle with a side length of (No Response) 8 along the x -axis and a side length of (No Response) 16 along the y -axis. (Give the numeric values.) (b) Master It The graph of g ( x ) consists of two straight lines and a semicircle. Use it to evaluate the integral. Step 1 of 3 On the interval the graph is a semi-circle. Since the semi-circle is below the x -axis, then is the negative of the area of this semi-circle. The radius of this semi-circle is (Give the numeric value.) g ( x ) dx b a g ( x ) dx 8 0 g ( x ) dx 30 10 g ( x ) dx 24 8 [ 8 , 24 ], g ( x ) dx 24 8 r = (No Response) 8 .
11/17/14, 2:15 PM Math 125 HW_1C Page 7 of 15 (c)

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