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# Calculus I MA 111

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This 8 page Class Notes was uploaded by Eldon Trantow on Monday October 19, 2015. The Class Notes belongs to MA 111 at Rose-Hulman Institute of Technology taught by Staff in Fall. Since its upload, it has received 16 views. For similar materials see /class/225075/ma-111-rose-hulman-institute-of-technology in Mathematics (M) at Rose-Hulman Institute of Technology.

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Date Created: 10/19/15

MA 111 Differentiation Quiz Solutions 2 Fall 2003 2004 Instructions Do all ten problems without the aid of any books notes calculators or computers No partial credit will be given Follow instructions and show appropriate work 1 y x713 413 y x7134c3 4x474x33x73 d y 16x3712x23 d1 2 Put the derivative in factored form after completing the di erentiation y 6 20 cos35L 7 627 sin3x dy d 7267290 cos35L 7 367290 sin3x 7 7267270 sin3x 367270 c0s3x c 7567290 cos35L 7 627 sin3x 76 2m5 c0s3x sin3x 03 y tan2c sec3x d d y 2 sec22x sec3x 3 tan2x sec3x tan35L CL 4 Put the result over a common denominator in hx4 7 h h2 dw 1 71 2b 47hh2 h db 2 V4 7 h h2 4 7 h h2 7h 2h2 x47hh22 47hh2 872h2h27h2h2 2x47hh2 873h4h2 2x47hh2 q D w arcsin2x 7 3 7x3x76 dy d1 x3x767 x753x21 W 72x3 15582 71 m y 2x3ln2x372x dy 2 21 2 3 7 2 3 2 cm 11 23 21n2x3 w arctan dyi 15 7 5 dx712725x2 x33xlny2y20 y 3x231ny3xg4yy 0 34yy 7358731113 y 73x273lny 3 4y 73x2y73ylny 3x 4312 10 y 57571 5 e w 7 lt60 6 0 ex w 7 gm am am 67m dcc 50 6 2 600 6790f 69 7 6790f 500 6 2 600 6700f ex 7 emf 50 6 2 62m 2 57290 7 62m 7 2 672T 500 6 2 MA 111 Calculus l Riemann Sums In this note We nd the various sums for the area under the curve 17 12 and between at a 0 and at Recall that a Riemann sum is given by n 3mm 1 or where n is the number of intervals7 at is chosen to lie in the i th interval 171 S a S 1n and A1 at 7 11 is the Width of the i th interval For all our examples Will make all the A1 the same7 name y 7 a Ax Aar n The Riemann sum then becomes n b 7 a n A l 12m n gm Note that in the text the points 6 772 and M are sometimes used instead of at Left hand sums First a picture left hand sums As mentioned in Class the left hand end of the i th interval is a 7 1A5L Thus AZfa 71A5L i1 For our particular function fa 7 1A5L 1 7 Thus our Riemann sum is 2 1 quot 121 1 quot 13927 2i1 17 17 n n n2 n2 11 11 We compute 2 nn12n1 nn1 i n7 6H3 n 7713 2n33n2n n2n n 7 6n3 n3 n3 717171711171 i 3 2n 6712 n n2 n2 72171 73 2n 6H2 723n71 73 6H2 Right hand sums Again a picture right hand sums The right hand end of the i th interval is a iAap Thus imam Axifm may 1 1 For our particular function fa iAar 1 7 Thus our Riemann sum is 2 n 2 1 lt1 1 gt 1 1 L2 72 l n n l n We compute 1 72 7272 12n 1 E 7 a 6n 72na3n2n 6723 1li 1 3 272 6722 2 1 1 3 272 6722 72 3n1 73 6722 Midpoint sums Again a picture Midpoint sums 2 1A1 Thus The midpoint of the i th interval is a Zmrmx 11 Axfa 21 1A1 For our particular function fa 7 1 7 271 sum is 2 2quot Thus our Riemann 1quot 223971 2 1quot 42274241 1 1 n lt 272 gt n g 4722 We compute 1quot ohm11quot 4quot 4quot1quot ggyiw ggymzx mzwm h 12 1 inm 3n 7272 1 n 2723 n 672 W 2na3n2n n2n 1 6723 2723 4722 1L 7272 6722 272 2722 4722 3W Comparison and upper sums and lower sums Observe that the limit of each of the three approximating sums is g and so it make sense to de ne the area to be this limit Since the function is decreasing the upper sums equal the left hand sums and the lower sums equal the left hand sums ie 2 37171 US 3 6712 7 2 3nl LS 7 3 6712 Thus the equation LSgAgUS becomes 2 3nllt2lt23n71 3 6712 7373 6712 which is clearly true

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