MATH 132 EXAM I Solutions FALL 2010 1 Estimate the area under the graph of 0 B B from x to x using 4 approximating rectangles and midpoints i e using M E F G H I J K L M N Solution B Q 0 0 0 0 F 2 Which of the following limits represents the area under 0 B B10 10 lim 3 8 8 8 3 8 lim 82 2 283 10 8 3 8 10 lim 3 8 8 8 3 8 lim 8 3 8 8 3 8 E H K N I L 2 B 4 10 3 82 10 G lim 8 2 23 8 8 8 3 8 3 8 8 10 lim 8 3 J lim 283 8 8 8 8 3 3 8 8 lim 3 10 M lim 8 283 10 8 8 3 8 8 3 F lim 8 8 Solution For V8 B 8 8 The first term in the summation will be 8 D is the one that satisfies both properties 3 8 6 8 8 3 1 lim 3 8 3 8 represents which of the following definite integrals 6 E B B B F B B G 3 x B B H B B B I B B B J B B B K B B B L B B B Solution B 8 8 9 6 The first term in the summation 3 8 8 J 9 and 0 B B B we get I satisfying both None of the others work 4 0 B B lim 8 8 0 B3 where the term inside the square bracket is R8 8 3 the Riemann sum with B3 being the right endpoint Using the formula 3 8 8 8 8 we have B B lim R8 where R8 8 3 E 8 8 8 F 8 8 8 G 8 8 8 H 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 I 10 J 8 K 8 8 8 8 8 8 8 8 8 8 8 8 L 6 M 8 N 10 8 8 8 8 3 3 0 B3 Solution 8 8 and B3 8 So 0 B3 8 Then 3 8 8 8 8 8 8 0 B3 3 8 Hence R8 8 8 8 3 3 3 8 8 8 8 8 8 8 8 K 8 8 8 6 5 Evaluate the integral l 2B l B by interpreting it in terms of areas E 4 F 6 G 8 H 10 I 12 J 14 K 16 L 18 M 20 N 22 Solution B B B B M 6 Find a particular solution for the indefinite integral sec sec tan E F 8 G 8 H 2 I sec J 8 K 8 L 8 M 8 N 8 Solution sec sec tan tan sec G G B B 7 If 2 B then find 2w E F G H I J K L M N Solution By Fund Theorem 2w B B B B Then 2w L 1 4 8 Using the substitution 1 tan t write the integral as an integral in the variable u 1 A F G H I 1 L M N Solution 9 and 1 4 1 Then K cos 8 1 cos 8 J 1 K 3 1 4 9 Evaluate the integral 0 9 1 1 1 1 1 E F G H I J 1 K 1 L 1 M 1 N 1 1 4 1 4 Solution 0 9 0 9 38 l 1 1 E 10 Find the total area between the curve 0 B x B and the x axis on 0 3 Recall The total area is area below the x axis area above the x axis E F G H I J K L M N Solution The area below the x axis is for 0 x 2 That area is B B B B B B The area above the x axis is for x That area is B B B Therefore the total area L 11 Evaluate the integral 1 to 4 decimal places E 0 2576 F 0 3896 G 0 4765 H 0 5143 I 0 5896 J 0 6854 K 0 7453 L 0 79 M 0 8642 N 0 9468 Solution 1 The first is arcsin t 1 The second is by the substitution Then l Then 1 1 F 12 Evaluate the integral B4 B B where and b are positive constants A F G D I J K L M N Solution Substitution B B B Then B4 B B I l 13 On what interval is the curve C 0 B concave upward E F G H I J K L M N B B B ww Solution By Fund Theorem 0 w B B By quotient rule f B B f ww B for B so that is where f x is concave upward E 4 14 Use integration by parts to evaluate the integral 8 E 1 F 1 68 G 1 68 H 1 68 I 68 1 J 68 1 K 68 1 L 68 1 M 1 N 68 1 8 Solution arctan t So 8 8 8 68 X 2 8 8 68 l 1 68 1 68 H Use the substitution B and then integration by parts to evaluate 1 38 0 E 1 K 1 F 1 G 1 L M 1 1 H N Solution i Substitution B B 1 I 1 1 J 1 B B B 1 1 Then 0 38 B 38 B B 33 Next integration by parts B 38 B B B 9 B So B 38 B B B 9 B 9 B B B 9 B 38 B 1 1 Then B 38 B B B 9 B 38 B l 1 L e 16 Evaluate the integral 1 B3 ln B B Write the answer to one decimal place E F G H I J K L M N Solution Integration by parts 68 B B B B B So B3 ln B B B 68 B B B 68 B B B Then e B3 ln B B 1 68 B B B l F 5 17 Using substitution find the general solution for the following two indefinite integrals In each case clearly state the substitution and get the final answer in terms of the variable x 5 points each 38 B 9 B B 9 B 38 B Solution 38 B 9 B B 9 B B The substitution is 9 B 38 B 9 B B Then we get 68l l G 68 9 B G x x 5 B Solution Substitution B B B Then x x 5 B G B B G Find a general solution for the indefinite integral using the following two steps each step is worth 5 points Find a substitution B 0 which allows you to rewite the integral as B B B for some constant c for some function g A Solution x x From this we …
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