# 6.3_MAT_266_ONLINE_B_Spring_2020..Section_6.3.pdf - Larena...

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Larena Whitney-krug Boerner MAT 266 ONLINE B Spring 2020 Assignment Section 6.3 due 03/26/2020 at 11:59pm MST 1. (1 point) Evaluate the integral. Z 6 5 7 x - 20 x 2 - 5 x + 6 dx Solution: SOLUTION We use the method of partial fractions to first simplify the inte- grand. The denominator of 7 x - 20 x 2 - 5 x + 6 can be factored as x 2 - 5 x + 6 = ( x - 3 )( x - 2 ) . So the partial fraction decomposition of 7 x - 20 x 2 - 5 x + 6 is of the form A x - 3 + B x - 2 , where we must solve for the A and B so that we have equality: 7 x - 20 x 2 - 5 x + 6 = A x - 3 + B x - 2 7 x - 20 x 2 - 5 x + 6 = A ( x - 2 ) ( x - 3 )( x - 2 ) + B ( x - 3 ) ( x - 3 )( x - 2 ) 7 x - 20 x 2 - 5 x + 6 = A ( x - 2 )+ B ( x - 3 ) x 2 - 5 x + 6 Therefore we must have the numerators equal: 7 x - 20 = A ( x - 2 )+ B ( x - 3 ) . Since the above must hold for all x , we see that if we let x = 3 then 7 ( 3 ) - 20 = A ( 3 - 2 )+ 0 = 1 = 1 A = A = 1. Similarly, if we let x = 2, then 7 ( 2 ) - 20 = 0 + B ( 2 - 3 )+ 0 = - 6 = - 1 B = B = 6. So 7 x - 20 x 2 - 5 x + 6 = 1 x - 3 + 6 x - 2 and Z 6 5 7 x - 20 x 2 - 5 x + 6 dx = Z 6 5 ( 1 x - 3 + 6 x - 2 ) dx = h 1ln x - 3 + 6ln x - 2 i 6 5 = h 1ln 3 + 6ln 4 i - h 1ln 2 + 6ln 3 i = 1ln 3 2 + 6ln 4 3 Correct Answers: 2.13155754281885 2. (1 point) Evaluate the indefinite integral. Z x 3 + 6 x 2 + 3 x + 2 dx Answer = + C Solution: SOLUTION We need long division of the integrand, x 3 + 6 x 2 + 3 x + 2 = x 3 + 0 x 2 + 0 x + 6 x 2 + 3 x + 2 : We have x - 3 R 7 x + 12 x 2 + 3 x + 2 x 3 + 0 x 2 + 0 x + 6 x 3 + 3 x 2 + 2 x - 3 x 2 - 2 x + 6 - 3 x 2 - 9 x - 6 7 x + 12 So x 3 + 6 x 2 + 3 x + 2 = x - 3 + 7 x + 12 x 2 + 3 x + 2 , We now use the method of partial fractions on 7 x + 12 x 2 + 3 x + 2 . The denominator of 7 x + 12 x 2 + 3 x + 2 can be factored as x 2 + 3 x + 2 = ( x + 2 )( x + 1 ) . So the partial fraction decomposition of 7 x + 12 x 2 + 3 x + 2 is of the form A x + 2 + B x + 1 , where we must solve for the A and B so that we have equality: 7 x + 12 x 2 + 3 x + 2 = A x + 2 + B x + 1 7 x + 12 x 2 + 3 x + 2 = A ( x + 1 ) ( x + 2 )( x + 1 ) + B ( x + 2 ) ( x + 2 )( x + 1 ) 7 x + 12 x 2 + 3 x + 2 = A ( x + 1 )+ B ( x + 2 ) x 2 + 3 x + 2 Therefore we must have the numerators equal: 7 x + 12 = A ( x + 1 )+ B ( x + 2 ) . Since the above must hold for all x , we see that if we let x = - 2 then 7 ( - 2 )+ 12 = A ( - 2 + 1 )+ 0 = ⇒ - 2 = - 1 A = A = 2. Similarly, if we let x = - 1, then 7 ( - 1 )+ 12 = 0 + B ( - 1 + 2 )+ 0 = 5 = 1 B = B = 5. So 7 x + 12 x 2 + 3 x + 2 = 2 x + 2 + 5 x + 1 And so Z x 3 + 6 x 2 + 3 x + 2 dx = Z x - 3 + 7 x + 12 x 2 + 3 x + 2 dx Z x - 3 + 2 x + 2 + 5 x + 1 dx = 1 2 x 2 - 3 x + 2ln x + 2 + 5ln x + 1 + C Correct Answers: (x**2)/2 + -3*x + 2*ln(abs(x+2))+5*ln(abs(x+1)) 1
3. (1 point) Note: You can get full credit for this problem by just entering the final answer (to the last question) correctly. The initial questions are meant as hints towards the final answer and also allow you the opportunity to get partial credit. Consider the indefinite integral Z x 2 ( x - 3 )( x - 4 ) 2 dx Then the integrand has partial fractions decomposition a x - 3 + b x - 4 + c ( x - 4 ) 2 where a = b = c = Integrating term by term, we obtain that Z x 2 ( x - 3 )( x - 4 ) 2 dx = + C Solution: SOLUTION We use the method of partial fractions to first simplify the inte- grand. The partial fraction decomposition of x 2 ( x - 3 )( x - 4 ) 2 is of the form a x - 3 + b x - 4 + c ( x - 4 ) 2 because we have a linear factor, ( x - 3 ) that is not repeated, and a linear factor ( x - 4 ) that is repeated twice.