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Objectives Use pattern recognition to find an indefinite integral. Use a change of variables to find an indefinite integral. Use the General Power Rule for Integration to find an indefinite integral. ANTIDIFFERENTIATION OF A COMPOSITE FUNCTION Let g be a function whose range is an interval I and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then f ( g ( x)) g '( x)dx F ( g ( x)) c If u = g(x), then du = g’(x)dx and f (u )du F (u ) c PATTERN RECOGNITION In this section you will study techniques for integrating composite functions. This is split into two parts, pattern recognition and change of variables. u-substitution is similar to the techniques used for the chain rule in differentiation. PATTERN RECOGNITION FOR FINDING THE ANTIDERIVATIVE Find 5cos 5xdx Let g(x)=5x and we have g’(x)=5dx So we have f(g(x))=f(5x)=cos5x From this, you can recognize that the integrand follows the f(g(x))g’(x) pattern. Using the trig integration rule, we get cos 5x 5dx sin 5x c You can check this by differentiating the answer to obtain the original integrand. RECOGNIZING PATTERNS Look at the more complex part of the function- the stuff ‘inside’. Does the stuff outside look like the derivative of the stuff inside? 2 4 2 2 x ( x 1) dx u x 1 du 2 x x 2 1 5 5 c RECOGNIZING PATTERNS 3x 2 x3 1 dx u x3 1 du 3x 2 3/ 2 2 3 x 1 c 3 RECOGNIZING PATTERNS 2 2 sec x tan x 3 dx u tan x 3 du sec x tan x 3 2 2 c CHANGE OF VARIABLES Find 5cos 5xdx Let u=5x and we have du=5dx cos udu sin u c cos 5x 5dx sin 5x c You can check this by differentiating the answer to obtain the original integrand. CHANGE OF VARIABLES A.K.A. U-SUBSTITUTION 2x x Example 3 2 1 dx 2 u x 1 du 2 xdx 2 2x x u du 2 x 2 1 dx 2 u 2 1 2 1 2 1 3 c 3 c u 3 3 c PATTERN RECOGNITION The integrands in Example 1 fit the f(g(x))g'(x) pattern exactly—you only had to recognize the pattern. You can extend this technique considerably with the Constant Multiple Rule Many integrands contain the essential part (the variable part) of g'(x) but are missing a constant multiple. In such cases, you can multiply and divide by the necessary constant multiple, as shown in Example 3. MULTIPLYING AND DIVIDING BY A CONSTANT Example 3 xx 2 1 dx 2 u x 1 du 2 xdx 2 2 1 2 2 x x 1 dx 2 1 2 1 3 u du u c 2 6 3 1 2 now back substitute : x 1 c 6 MULTIPLYING AND DIVIDING BY A CONSTANT ALTERNATE METHOD Example 3 xx 2 1 dx 2 let u x 2 1 du 2 xdx but we have only x, not 2 x...so we do algebra 1 du xdx now substitute : 2 1 u 2 du 2 1 2 1 3 u du u c 2 6 3 1 2 now back substitute : x 1 c 6 CHANGE OF VARIABLES EXAMPLE 4 2 x 1dx EXAMPLE 4 – CHANGE OF VARIABLES Find Solution: First, let u be the inner function, u = 2x – 1. Then calculate the differential du to be du = 2dx. Now, using and substitute to obtain EXAMPLE 4 – SOLUTION cont’d CHANGE OF VARIABLES Example 5 x 2 x 1dx CHANGE OF VARIABLES Example 5 x 2 x 1dx let _ u 2 x 1 du 2dx solve for x : (u 1) / 2 1 u 1 2 du Substitute : x 2 x 1dx u 2 2 1 3 1 1 1 1 2 5/ 2 2 3/ 2 2 2 2 u 1u du u u du u u c 4 4 45 3 5 3 1 1 Back _ Substitute : 2 x 1 2 2 x 1 2 c 10 6 CHANGE OF VARIABLES EXAMPLE Example 6 2 sin 3x cos 3xdx EXAMPLE 7 – SUBSTITUTION AND THE GENERAL POWER RULE EXAMPLE 7 – SUBSTITUTION AND THE GENERAL POWER RULE cont’d GUIDELINES FOR MAKING A CHANGE OF VARIABLES 1. 2. 3. 4. 5. 6. Choose a substitution u=g(x). Usually it is best to choose the INNER part of a composite function, such as a quantity raised to a power. Compute du=g’(x)dx Rewrite the integral in terms of the variable u Find the resulting integral in terms of u Replace u by g(x) to obtain an antiderivative in terms of x. Check your answer by differentiating. GENERAL POWER RULE FOR INTEGRATION Theorem: The General Power Rule for Integration If g is a differentiable function of x, then g x g x g x dx n 1 n 1 n Equivalently, if u=g(x), then u n 1 u du n 1 c, n 1 n c, n 1 CHANGE OF VARIABLES FOR DEFINITE INTEGRALS If the function u=g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then b g b a g a f g x g x dx f u du Basically, this gives us permission to find the area under the curve with these types of integration problems. DEFINITE INTEGRALS AND CHANGE OF VARIABLES Example 8 xx 1 0 2 1 dx 3 EXAMPLE 8 – CHANGE OF VARIABLES Evaluate Solution: To evaluate this integral, let u = x2 + 1. Then, you obtain Before substituting, determine the new upper and lower limits of integration. EXAMPLE 8 – SOLUTION Now, you can substitute to obtain cont’d EXAMPLE 8 – SOLUTION cont’d Try rewriting the antiderivative in terms of the variable x and evaluate the definite integral at the original limits of integration, as shown. Notice that you obtain the same result. DEFINITE INTEGRALS AND CHANGE OF VARIABLES Example 9: This problem requires a little more work: 5 x 1 2 x 1 dx INTEGRATION OF EVEN AND ODD FUNCTIONS Even with a change of variables, integration can be difficult. Occasionally, you can simplify the evaluation of a definite integral over an interval that is symmetric about the y-axis or about the origin by recognizing the integrand to be an even or odd function (see Figure 4.40). Figure 4.40 INTEGRATION OF EVEN AND ODD FUNCTIONS EXAMPLE 10 – INTEGRATION OF AN ODD FUNCTION Evaluate Solution: Letting f(x) = sin3x cos x + sin x cos x produces f(–x) = sin3(–x) cos (–x) + sin (–x) cos (–x) = –sin3x cos x – sin x cos x = –f(x). EXAMPLE 10 – SOLUTION cont’d So, f is an odd function, and because f is symmetric about the origin over [–π/2, π/2], you can apply Theorem 4.16 to conclude that Figure 4.41 HOMEWORK: p.304 1,3,9,14,16,20,25,27,31,33,34,41,45,51,54,59,69, 79,81,95,103 These are the minimum to help you PASS this next quiz/test. I suggest you complete all problems in the sections relating to what was covered in the notes. Students taking the AP Exam in the spring should also look at 107, 108, 111-112, and the word problems (see me for assistance when you get to them).