Barrons AP Calculus - David Bock

Using v(1) = 6, we get 6 = 2(1)^2 − 3(1) + C 1 , and C 1 = 7, from which it follows that v(t) = 2t^2

− 3t + 7. Since

Using f (2) = 5, we get + 14 + C 2 , so Thus,

For more examples of motion along a line, see Chapter 8, Further Applications of Integration, and

Chapter 9, Differential Equations.

Chapter Summary

In this chapter, we have reviewed basic skills for finding indefinite integrals. We’ve looked at the
antiderivative formulas for all of the basic functions and reviewed techniques for finding
antiderivatives of other functions.
We’ve also reviewed the more advanced techniques of integration by partial fractions and
integration by parts, both topics only for the BC Calculus course.

Practice Exercises

Directions: Answer these questions without using your calculator.

1.

(A) x^3 − x^2 + C

(B) 3 x^3 − x^2 + 3x + C

(C) x^3 − x^2 + 3x + C

(D)

(E) none of these

2.

(A)