England_MAT_265_ONLINE_B_Summer_2019.jlalonzo.Section_4.7.pdf - Jesse Alonzo Assignment Section 4.7 due at 11:59pm MST England MAT 265 ONLINE B Summer

England_MAT_265_ONLINE_B_Summer_2019.jlalonzo.Section_4.7.pdf

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Jesse Alonzo England MAT 265 ONLINE B Summer 2019 Assignment Section 4.7 due 08/09/2019 at 11:59pm MST 1. (1 point) Find the antiderivatives for dy du = 3 u 5 - 7 u 2 - 4 . y = + C . Solution: If dy du = 3 u 5 - 7 u 2 - 4, then y ( u ) = 3 u 5 + 1 5 + 1 - 7 u 2 + 1 2 + 1 - 4 u + C = 1 2 u 6 - 7 3 u 3 - 4 u + C Answer(s) submitted: uˆ6/2 - 7uˆ3/3 - 4u (correct) Correct Answers: (3*u**(5+1))/(5+1) - (7*u**(2+1))/(2+1) - 4*u 2. (1 point) Find the antiderivatives for dy dx = 8 e x + 5 . y = + C . Solution: If dy dx = 8 e x + 5, then y ( x ) = 8 e x + 5 x + C Answer(s) submitted: 8eˆx+5x (correct) Correct Answers: 8*exp(x) + 5*x 3. (1 point) Consider the function f ( x ) = 20 x 3 - 6 x 2 + 18 x - 8. Enter an antiderivative of f ( x ) Solution: The general antiderivative of f ( x ) is F ( x ) = 20 x 4 4 - 6 x 3 3 + 18 x 2 2 - 8 x = 5 x 4 - 2 x 3 + 9 x 2 - 8 x + C An antiderivative can be obtained by assigning any arbitrary value to C . For instance, for C = 0 we obtain the antideriva- tive F ( x ) = 5 x 4 - 2 x 3 + 9 x 2 - 8 x Answer(s) submitted: 5xˆ4-2xˆ3+9xˆ2-8x (correct) Correct Answers: 5*xˆ4-2*xˆ3+9*xˆ2-8*x 4. (1 point) Find the most general antiderivative for the function 1 4 u . Note: Don’t enter the +C . It’s included for you. Antiderivative = + C . Solution: SOLUTION We rewrite the funtion in terms of power of u 1 4 u = 1 4 u - 1 / 2 Thus the most general antideriative is 1 4 u 1 / 2 1 / 2 + C = 1 2 u 1 / 2 + C Answer(s) submitted: (1/2) ((u)ˆ(1/2)) (correct) Correct Answers: (2/4)*u**(1/2) 5. (1 point) Find the most general antiderivative for the function 6 x 4 - 6 x 5 - 3 . Note: Don’t enter the +C . It’s included for you. Antiderivative = + C . Solution: SOLUTION We rewrite the function in terms of powers of x : 6 x 4 - 6 x 5 - 3 = 6 x 4 - 6 x - 5 - 3 The most general antiderivative is: 6 x 5 5 - 6 x - 5 + 1 - 5 + 1 - 3 x + C = 6 5 x 5 + 3 2 x - 4 - 3 x + C Answer(s) submitted: (6xˆ5) / (5) + (3)/(2xˆ4)-(3x) (correct) Correct Answers: 6*(x**5)/5 - 6*(x**(-5+1))/(-5+1) - 3*x 1
6. (1 point) Let f ( x ) = 10 x - 8 e x . Enter an antiderivative of f ( x ) Solution: The general antiderivative of f ( x ) is F ( x ) = 10ln | x |- 8 e x + C Since we are asked for an antiderivative, we can choose any value for the arbitrary constant C . In particular, for C = 0 we have F ( x ) = 10ln ( | x | ) - 8 e x Note that any other value for the constant C would give a correct answer. Answer(s) submitted: 10ln(abs(x))-8eˆx (correct) Correct Answers: 10 * ln(abs(x)) -8 * eˆx 7. (1 point) Find the antiderivatives for dx dt = 2 t - 1 + 5 . x = + C . Solution: The general antiderivative of dx dt = 2 t - 1 + 5 , t 6 = 0 is x ( t ) = 2ln | t | + 5 t + C Answer(s) submitted: 2ln(abs(t)) + 5t (correct) Correct Answers: 2*ln(abs(t)) + 5*t 8. (1 point) Let f ( x ) = - 13 1 - x 2 .

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