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GROUP 6’s presentation on ASSIGNMENT # 5 AREAS BY DEFINITE INTEGRATION p. 335 # 16

GROUP 6’s presentation on ASSIGNMENT # 5 AREAS BY DEFINITE INTEGRATION p. 335 # 16. AREAS BY DEFINITE INTEGRATION. Given : y = 2 e x ; y = (2x / ln 2) + 2 Using a simultaneous solution, we will get the points at which they intersect: y = 2 e x -y = - (2x / ln 2) - 2

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GROUP 6’s presentation on ASSIGNMENT # 5 AREAS BY DEFINITE INTEGRATION p. 335 # 16

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  1. GROUP 6’spresentation onASSIGNMENT # 5AREAS BY DEFINITE INTEGRATIONp. 335 # 16

  2. AREAS BY DEFINITE INTEGRATION • Given : y = 2 ex ; y = (2x / ln 2) + 2 • Using a simultaneous solution, we will get the points at which they intersect: y = 2 ex -y = - (2x / ln 2) - 2 2 ex - ( 2x / ln2) - 2 = 0 2 ex - ( 2x / ln2) = 2 [ 2 ex - ( 2x / ln2) = 2 ] ln

  3. AREAS BY DEFINITE INTEGRATION [ 2 ex - ( 2x / ln2) = 2 ] ln ln 2 + ln ex - (ln 2x - lnln 2) = ln2 x ln e - ln 2x + lnln2 = 0 x (1) - (ln 2 + ln x) + lnln2 = 0 x - ln x = ln 2 - lnln2 x = 0.693 y = 3.999 The equations intersect at (0.693,3.999) & (0, 2)

  4. AREAS BY DEFINITE INTEGRATION Thus, A =  [ ( 2x / ln 2) + 2 - 2 ex ] dx = (1 / ln 2) [ ( 2x2)/2 ] + 2x - 2 ex = [ (x2 / ln 2) + 2x - 2 ex ]0.6930 = -2 + 1.92 = - 0.07944 = | -0.07994 | A= 0.07994

  5. GRAPH AREA

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