Module 18 - Antiderivatives as Indefinite Integrals and Differential Equations

Module 18 - Answers

Lesson 1

Answer 1

18.1.1

Because , and , it appears that the exponent of x in the antiderivative is one greater than the exponent of the original function and that the antiderivative is divided by the value of its exponent. Predict that is .

Answer 2

18.1.2

Answer 3

18.1.3

The TI-89 results may appear to be different from your predictions, however they are algebraically equivalent to and , respectively.

Answer 4

18.1.4

This rule works for all values of n except n = -1. Recall that
was discussed in Lesson 16.2.

Lesson 2

Answer 1

18.2.1

then

Answer 2

18.2.2

The family of curves is shown in a [-4 , 4 ] x [-10, 10] window with
C = -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, and 10. The curve with C = 0 is darker.

Lesson 3

Answer 1

18.3.1

Answer 2

18.3.2

The general solution is y = Ce^x .

Answer 3

18.3.3

The solution to y' = 2x with y(2)=1 is y = x² – 3.

Answer 4

18.3.4

Lesson 4

Answer 1

18.4.1

y(3) = 6

Answer 2

18.4.2

Answer 3

18.4.3

Enter y1' = y1, not y1' = y, be sure to clear yi1, and use a [-3, 3] x [-5, 10] window.

Instead of parallel line segments in columns, line segments in rows are parallel.

Self Test

Answer 1

The indefinite integral , displayed in a [-2 , 2 ] x [-5, 5] window, with C = -6, -4, -2, 0, 2, 4, 6.

Answer 2

y = (x³ – 3x² + 6x – 6)e^x + C

Answer 3

Answer 4

y(3) = 3 ln(3)

3.29584

Answer 5

[-3, 3] x [-2, 2]

Answer 6

[-3, 3] x [-1, 3]