1. 1 § 12.1 Antiderivatives and Indefinite Integrals The student will learn about: antiderivatives, indefinite integrals, and applications.

2. 2 Antiderivatives. Many operations in mathematics have reverses – inverses – such as multiplication and division. The reverse operation of finding a derivative (antiderivative) will now command our attention.

3. 3 Examples Find a function that has a derivative of x. Find a function that has a derivative of 2x. x2x2 Find a function that has a derivative of x 2. x 2 /2 Oops! x 3 /3 x2x2 Since d/dx [x 2 ] = 2x Since d/dx [x 2 /2] = x Since d/dx [x 3 /3] = x 2

4. 4 Examples - continued The above functions that you found are antiderivatives. Find a function that has a derivative of 2x. x2x2 Note that you can find more than one such function? x 2 + 3x 2 - 5

5. 5 Theorem 1 If a function has more than one antiderivative then the antiderivatives differ by at most a constant.

6. 6 Antiderivatives and Indefinite Integrals – Algebraic Forms. Notation Let f (x) be a function, then the family of all functions that are antiderivatives of f (x) is called the indefinite integral and has the symbol ∫ f (x) dx OR F (x) + C ∫ f (x) dx = F (x) + C if F’ (x) = f (x). And The symbol ∫ is called an integral sign, and the function f (x) is called the integrand. The symbol dx indicates that the antidifferentiation is performed with respect to the variable x. the arbitrary constant C is called the constant of integration.

7. 7 Example Now, let’s get to the basics and make this easy! since

8. 8 Indefinite Integral Formulas and Properties. 1. Power rule. 2.∫ e x dx = e x + C. 3. ∫ 1/x dx = ln |x| + C, x ≠ 0. ∫ x - 1 dx = 4.∫ k f (x) dx = k ∫ f (x) dx 5.∫ [f (x) ± g (x)] dx = ∫ f (x) dx ± ∫ g (x) dx. Integral of a sum or difference.

9. 9 Examples c. ∫ 5 x - 3 dx = a. ∫ 444 dx = b. ∫ x 3 dx = e. ∫ (x 4 + x + x ½ + 1 + x – ½ + x – 2 ) dx = d. ∫ x 2/3 dx = x 4 /4 + CPower rule. 444x + CIntegral of a constant. 5 x -2 /-2 + CPower rule. Power rulex 5/3 / 5/3 + C =3/5 x 5/3 + C x 5 /5 + x 2 /2 + 2 x 3/2 /3 + x + 2 x 1/2 – x – 1 + C f. ∫ x - 1 dx = x 0 /0 + C = Undefined !!

10. 10 Examples g. ∫ 4 e x dx = h. ∫ 4/x dx =4 ln |x| + C 4e x + C

11. 11 Applications In spite of the prediction of a paperless computerized office, paper and paperboard production in the United States has steadily increased. In 1990 the production was 80.3 million short tons, and since 1970 production has been growing at a rate given by f ’ (x) = 0.048x + 0.95 Where x is years after 1970. Noting that f (20) = 80.3, find f (x). Continued on next slide.

12. 12 Applications - continued 0.024 x 2 + 0.95 x + c = f (x) Note, f ’ (x) = 0.048 x + 0.95 Where x is years after 1970. Noting that f (20) = 80.3, find f (x). 80.3 = (0.024)(20 2 ) + (0.95)(20) + c Noting that f (20) = 80.3 Continued on next slide. 80.3 = 28.6 + c and We need the integral of f ‘ (x) or ∫ (0.048x + 0.95) dx = c = 51.7 so f (x) = 0.024 x 2 + 0.95 x + 51.7 f (x) = 0.024 x 2 + 0.95 x + c

13. 13 Applications - continued f (0) = (0.024)(0 2 ) + (0.95)(0) + 51.7 = Note, f ’ (x) = 0.048t + 0.95 Where x is years after 1970. f (x) = 0.024 x 2 + 0.95 x + 51.7. f (30) = (0.024)(30 2 ) + (0.95)(30) + 51.7 = 51.7 101.8 Find f (0) and f (30), the production levels for 1970 and 2000. 51.7 short tons in 1970 101.8 short tons in 2000

14. 14 Summary. 2. 6. ∫ 1/x dx = ln |x| + C, x ≠ 0. 1. ∫ k dx = kx + C. 5. ∫ e x dx = e x + C. 3. ∫ k f (x) dx = k ∫ f (x) dx 4. ∫ [f (x) ± g (x)] dx = ∫ f (x) dx ± ∫ g (x) dx.

15. 15 Practice Problems §6.1; 1, 5, 9, 13, 17, 21, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 65, 69, 73, 77, 81, 83, 87, 89, 99, 103, 109, 111.