England_MAT_265_ONLINE_A_Spring_2020.Section_4.7.pdf - England MAT 265 ONLINE A Spring 2020 Assignment Section 4.7 4(1 point 1(1 point Find the

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England MAT 265 ONLINE A Spring 2020 Assignment Section 4.7 1. (1 point) Find the antiderivatives for dy du = 7 u 4 - 6 u 2 - 4 . y = + C . Solution: If dy du = 7 u 4 - 6 u 2 - 4, then y ( u ) = 7 u 4 + 1 4 + 1 - 6 u 2 + 1 2 + 1 - 4 u + C = 7 5 u 5 - 2 u 3 - 4 u + C Correct Answers: (7*u**(4+1))/(4+1) - (6*u**(2+1))/(2+1) - 4*u 2. (1 point) Find the antiderivatives for dy dx = 8 e x + 6 . y = + C . Solution: If dy dx = 8 e x + 6, then y ( x ) = 8 e x + 6 x + C Correct Answers: 8*exp(x) + 6*x 3. (1 point) Consider the function f ( x ) = 40 x 3 - 21 x 2 + 8 x - 2. Enter an antiderivative of f ( x ) Solution: The general antiderivative of f ( x ) is F ( x ) = 40 x 4 4 - 21 x 3 3 + 8 x 2 2 - 2 x = 10 x 4 - 7 x 3 + 4 x 2 - 2 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 ) = 10 x 4 - 7 x 3 + 4 x 2 - 2 x Correct Answers: 10*xˆ4-7*xˆ3+4*xˆ2-2*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 Correct Answers: (2/4)*u**(1/2) 5. (1 point) Find the most general antiderivative for the function 8 x 4 - 6 x 4 - 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 : 8 x 4 - 6 x 4 - 3 = 8 x 4 - 6 x - 4 - 3 The most general antiderivative is: 8 x 5 5 - 6 x - 4 + 1 - 4 + 1 - 3 x + C = 8 5 x 5 + 2 x - 3 - 3 x + C Correct Answers: 8*(x**5)/5 - 6*(x**(-4+1))/(-4+1) - 3*x 6. (1 point) Let f ( x ) = 9 x - 8 e x . Enter an antiderivative of f ( x ) Solution: The general antiderivative of f ( x ) is F ( x ) = 9ln | 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 ) = 9ln ( | x | ) - 8 e x Note that any other value for the constant C would give a correct answer. Correct Answers: 9 * ln(abs(x)) -8 * eˆx 1
7. (1 point) Find the antiderivatives for dx dt = 4 t - 1 + 3 . x = + C . Solution: The general antiderivative of dx dt = 4 t - 1 + 3 , t 6 = 0 is x ( t ) = 4ln | t | + 3 t + C Correct Answers: 4*ln(abs(t)) + 3*t 8. (1 point) Let f ( x ) = 14 1 - x 2 . Enter an antiderivative of f ( x ) + C Solution: The general antiderivative of f ( x ) is F ( x ) = 14arcsin ( 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 ) = 14arcsin ( x ) Note that any other value for the constant C would give a correct answer.
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