2. INTEGRATION ANTIDERIVATIVE: If F ' ( x ) = f ( x ), then F ( x ) is an antiderivative of f ( x ). If F ( x ) and G ( x ) are both antiderivatives of a function f ( x ) on the interval, then there is a constant C such that F ( x ) – G ( x ) = C.C. (Two antiderivatives of a function can only differ by a constant.) The arbitrary real number C is called an integration constant. INDEFINITE INTEGRAL If F ' ( x ) = f ( x ), then  f ( x ) dx = F ( x ) + C, for any real number C.C. POWER RULE: For any real number n  – 1, is the reciprocal of the exponent. Reciprocal means to turn upside down.

3. CONSTANT MULTIPLE RULE AND SUM OR DIFFERENCE RULE: If all indicated integrals exist,  k  f ( x ) dx = k   f ( x ), for any real number k, and  [ f ( x )  g ( x ) ] dx =  f ( x )   g ( x ). INDEFINITE INTEGRALS OF EXPONENTIAL FUNCTIONS:

4. INDEFINITE INTEGRAL OF x – 1 You need to study page 850 problems 5 – 36, 41, 42, and 45 – 52. Example 1: Determine the exponent first, then write its reciprocal in front of the variable. Example 2:

5. SUBSTITUTION METHOD : In general, for the types of problems we are concerned with, there are three cases. We choose u to be one of the following: 1. the quantity under a root or raised to a power; 2. the exponent on e;e; 3. the quantity in the denominator. Remember that some integrands may need to be rearranged to fit one of these cases. Example 3: Let u = x 2 + 5 Then du = 2x 2x dx du and dx are called differentials not derivatives. You must determine the differential du, you do not copy it from the problem. If you copy what you think du is from the problem most of the time you will be wrong. We need 2x 2x dx but we have too much. We have 10x dx, so rewrite the 10x as 5 ( 2x 2x ).

6. 5  u 6 du Make the substitution in the problem to simplify its looks. Replace the 2x dx with du and the ( x 2 + 5 ) with u. But the original problem was not in terms of u, it was in terms of x.x. You must re-substitute to get the final answer.

7. Example 4: Let u = 4x 4x 3 + 3 Then du = 12x 2 dx We need 12 x 2 dx but the problem only contains 3 x 2 dx. If we multiply by 4, 4 ( 3 x 2 dx ), we would have the 12 x 2 dx. We can not just multiply by 4 because that changes the value of the problem. To compensate we will write the fraction ¼ in front of the  symbol. (¼) · 4 = 1. We have not changed the value of the problem, only its looks. Notice the difference between this problem and the previous problem. In the first problem we did not write the 5 as a fraction because we had too much. In the second problem the fraction ¼ was used to compensate for the 4 we used that was not in the original problem.

8. Example 5: Let u = 5 x 3 Then du = 15 x 2 dx Example 6: Let u = x 2 Then du = 2 x dx Example 7: Let u = x 2 + x Then du = ( 2 x + 1 ) dx The parentheses are necessary. Without them it is wrong.

9. Example 8: Let u = x 3 + 1 Then du = 3 x 2 dx Example 9: Let u = 6 + 5 t Then du = 5 dt Notice t is still in the problem because it was never removed by the substitutions. It may not be left in the problem because everything must be in terms of u’s. Using the fact that u = 6 + 5 t, we can solve this equation for t in terms of u and substitute this into the problem for t.t. u = 6 + 5 t u – 6 = 5 t

10. Rewrite the problem so that it is easier to understand. Dividing by 5 is the same as multiplying by the fraction ⅕. Then distribute the u 1/ 2 to the expression ( u – 6 ) by multiplication using the rules of exponents. Notice the fraction of ⅕ is written with both of the terms. Study problems on page 859.