INTEGRATION BY PARTS

1. INTEGRATION BY PARTS

2. INTEGRATION BY PARTS ( Chapter 16 ) If u and v are differentiable functions, then ∫udv = uv – ∫vdu. There are two ways to integrate by parts; the standard method (above) and column integration. Both methods will be shown but on the test I will only ask for the column integration procedure. First write the expression as a product of two functions u and dv in such a way that ∫dv can be found. Example: ∫ xe 6xdx Choose dv = e 6xdx and u = x. Then du = 1dx Same concept as used in Test 4. You must remember that Find v by integrating dv: v = ∫dv = ∫ e 6xdx = Substitute into the formula: ∫udv = uv – ∫vdu. Evaluating Substituting yields

3. CONDITIONS FOR INTEGRATION BY PARTS Integration by parts can be used only if the integrand satisfies the following conditions. 1. The integrand can be written as the product of two factors u and dv. 2. It is possible to integrate dv to get v and to differentiate u to get du. 3. The integral ∫ vdu can be found. COLUMN INTEGRATION Is a technique that is equivalent to integration by parts. Create two columns, one entitled D ( for derivative) and the other one entitled I ( for integral ). Under the D write the part of the problem that will be differentiated. Under the I write the rest of the problem that will be integrated. Working with the column entitled D first, write the derivative of the function and on each subsequent line write the derivative of the preceding function until the answer of 0 is found. Next form the second column under I in a similar fashion except you will write an antiderivative of the preceding function until the second column has the same number of rows as the column entitled D.

4. Next, connect the terms diagonally with a line. Then alternately label each line with “+” then “–” until all lines are labeled. You may write the signs on each line or in front of each term of D that has been connected by a diagonal line. Starting with the first sign multiply the opposite ends of each diagonal line writing the answer as a sum or difference until all products have been formed remembering to write a + C at the end. D I EXAMPLE: ∫ 5x 3e 4xdx + 5x 3 Pay close attention to the order of the animation in the problem. It gives the order in which the solution is obtained. – 15x 2 + 30x – 30 0

5. Even though you might believe you should put the 16x 3 – 20x beneath the D, you place the Ln |7x|beneath D. The reason is we only know how to find its derivative. We do not know how to integrate it. EXAMPLE: ∫ (16x 3 – 20x)Ln |7x| dx D I + Ln |7x| 16x 3 – 20x Since this can never yield the value of zero by repeated derivatives we stop. – 4x 4 – 10x 2 Draw diagonal lines with alternating signs as before. Because there is no zero beneath D draw a horizontal line. The presence of a horizontal line indicates the product is to be integrated. + (4x 4 – 10x 2 )Ln|7x| – (4x 4 – 10x 2 )Ln |7x| – ∫ ( 4x 3 – 10x ) dx (4x 4 – 10x 2 )Ln |7x| – x 4 + 5x 2 + C Notice the sign has changed. Make certain you understand why.

6. Find the general solution and or particular solution for the differential equation. Algebraically separate the problem so that y’s are on one side of the equal sign and x’s are on the other side. Then integrate both sides remembering to write + C. See page 958 problems 1 – 30. Example 2: Example 1: In step 3, only a y 2 was factored out of the right expression even though all the numbers were divisible by three. We do not do complete factoring. We are only trying to separate the variables.

7. Example 3: Example 4: Cross multiplying gives Ln | y + 5 | = x + C

8. Find the particular solution for Substitute the given values of the variables and solve for C. Remember to write the final answer to the question. Answer:

9. Some problems are solved using a table of integrals. Page A – 4 in the back of your text book contains 17 examples. These are not the only examples, there are books dedicated to such examples. These solutions are beyond the scope of this course and may have taken years to solve. However people noticed that problems having the similar characteristics had similar solutions. What you do is find the example on page A – 4 that best matches your problem. You may have to do some algebraic work first. The table of integrals then shows how to write the answer to that example. Suppose we were trying to solve the following problem: Upon close inspection this problem is similar to example 15 on page A – 4: The a 2 does not mean we have to actually have the exponent two. In fact looking at the example’s solution indicates this is not important because the value of a is not needed. But if it was necessary, we could consider 7 as What is important is the sign between x 2 and the number is a plus sign.

10. Next, the example shows there is no number between the integral sign and the square root symbol. This can be remedied by rewriting the problem as In doing this, we must remember to multiply the entire answer by eight; not just part of it. The Table of Integrals states the solution to our problem is in the form of Making the indicated substitutions and remembering to multiply the entire answer by eight yields the following Work problems 23 – 28 on page 925. On the test I will tell you when to use the Table of Integrals.