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- Ch 7 Sec 7.4 Techniques of Integration James Stewart pdf online

Amit M.

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SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS |||| 473 39. (a) Use trigonometric substitution to verify that (b) Use the figure to give trigonometric interpretations of both terms on the right side of the equation in part (a). 40. The parabola divides the disk into two parts. Find the areas of both parts. 41. Find the area of the crescent-shaped region (called a lune) bounded by arcs of circles with radii and . (See the figure.) 42. A water storage tank has the shape of a cylinder with diam- eter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the total capacity is being used? 43. A torus is generated by rotating the circle about the -axis. Find the volume enclosed by the torus. xx 2 H11001 H20849y H11002 RH20850 2 H33527 r 2 R r Rr x 2 H11001 y 2 H33355 8y H33527 1 2 x 2 ¨ ¨ y=œ„„„„„a@-t@ t0 y a x y x 0 sa 2 H11002 t 2 dt H33527 1 2 a 2 sin H110021 H20849xH20862aH20850 H11001 1 2 xsa 2 H11002 x 2 35. Prove the formula for the area of a sector of a circle with radius and central angle . [Hint: Assume and place the center of the circle at the origin so it has the equation . Then is the sum of the area of the triangle and the area of the region in the figure.] ; 36. Evaluate the integral Graph the integrand and its indefinite integral on the same screen and check that your answer is reasonable. ; 37. Use a graph to approximate the roots of the equation . Then approximate the area bounded by the curve and the line . 38. A charged rod of length produces an electric field at point given by where is the charge density per unit length on the rod and is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field . 0 x y L P (a, b) EH20849PH20850 H9255 0 H9261 EH20849PH20850 H33527 y LH11002a H11002a H9261b 4H9266H9255 0 H20849x 2 H11001 b 2 H20850 3H208622 dx PH20849a, bH20850 L y H33527 2 H11002 xy H33527 x 2 s4 H11002 x 2 x 2 s4 H11002 x 2 H33527 2 H11002 x y dx x 4 sx 2 H11002 2 O x y RQ ¨ P PQRPOQ Ax 2 H11001 y 2 H33527 r 2 0 H11021 H9258 H11021 H9266H208622 H9258r A H33527 1 2 r 2 H9258 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS In this section we show how to integrate any rational function (a ratio of polynomials) by expressing it as a sum of simpler fractions, called partial fractions, that we already know how to integrate. To illustrate the method, observe that by taking the fractions and to a common denominator we obtain If we now reverse the procedure, we see how to integrate the function on the right side of 2 x H11002 1 H11002 1 x H11001 2 H33527 2H20849x H11001 2H20850 H11002 H20849x H11002 1H20850 H20849x H11002 1H20850H20849x H11001 2H20850 H33527 x H11001 5 x 2 H11001 x H11002 2 1H20862H20849x H11001 2H20850 2H20862H20849x H11002 1H20850 7.4 this equation: To see how the method of partial fractions works in general, let’s consider a rational function where and are polynomials. It’s possible to express as a sum of simpler fractions provided that the degree of is less than the degree of . Such a rational function is called proper. Recall that if where , then the degree of is and we write . If is improper, that is, , then we must take the preliminary step of dividing into (by long division) until a remainder is obtained such that . The division statement is where and are also polynomials. As the following example illustrates, sometimes this preliminary step is all that is required. EXAMPLE 1 Find . SOLUTION Since the degree of the numerator is greater than the degree of the denominator, we first perform the long division. This enables us to write M The next step is to factor the denominator as far as possible. It can be shown that any polynomial can be factored as a product of linear factors (of the form ) and irreducible quadratic factors (of the form , where ). For instance, if , we could factor it as The third step is to express the proper rational function (from Equation 1) as a sum of partial fractions of the form Ax H11001 B H20849ax 2 H11001 bx H11001 cH20850 j or A H20849ax H11001 bH20850 i RH20849xH20850H20862QH20849xH20850 QH20849xH20850 H33527 H20849x 2 H11002 4H20850H20849x 2 H11001 4H20850 H33527 H20849x H11002 2H20850H20849x H11001 2H20850H20849x 2 H11001 4H20850 QH20849xH20850 H33527 x 4 H11002 16 b 2 H11002 4ac H11021 0ax 2 H11001 bx H11001 c ax H11001 bQ QH20849xH20850 H33527 x 3 3 H11001 x 2 2 H11001 2x H11001 2 ln H11341 x H11002 1 H11341 H11001 C y x 3 H11001 x x H11002 1 dx H33527 y H20873 x 2 H11001 x H11001 2 H11001 2 x H11002 1 H20874 dx y x 3 H11001 x x H11002 1 dxV RS f H20849xH20850 H33527 PH20849xH20850 QH20849xH20850 H33527 SH20849xH20850 H11001 RH20849xH20850 QH20849xH20850 1 degH20849RH20850 H11021 degH20849QH20850 RH20849xH20850PQ degH20849PH20850 H33356 degH20849QH20850f degH20849PH20850 H33527 nnPa n HS33527 0 PH20849xH20850 H33527 a n x n H11001 a nH110021 x nH110021 H11001H11080H11080H11080H11001a 1 x H11001 a 0 QP fQP f H20849xH20850 H33527 PH20849xH20850 QH20849xH20850 H33527 2 ln H11341 x H11002 1 H11341 H11002 ln H11341 x H11001 2 H11341 H11001 C y x H11001 5 x 2 H11001 x H11002 2 dx H33527 y H20873 2 x H11002 1 H11002 1 x H11001 2 H20874 dx 474 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION x-1 ≈+x+2 ˛-≈ ≈+x ≈-x 2x 2x-2 2 ˛ +x ) A theorem in algebra guarantees that it is always possible to do this. We explain the details for the four cases that occur. CASE I N The denominator Q(x) is a product of distinct linear factors. This means that we can write where no factor is repeated (and no factor is a constant multiple of another). In this case the partial fraction theorem states that there exist constants such that These constants can be determined as in the following example. EXAMPLE 2 Evaluate . SOLUTION Since the degree of the numerator is less than the degree of the denominator, we don’t need to divide. We factor the denominator as Since the denominator has three distinct linear factors, the partial fraction decomposition of the integrand (2) has the form To determine the values of , , and , we multiply both sides of this equation by the product of the denominators, , obtaining Expanding the right side of Equation 4 and writing it in the standard form for polyno- mials, we get The polynomials in Equation 5 are identical, so their coefficients must be equal. The coefficient of on the right side, , must equal the coefficient of on the left side—namely, 1. Likewise, the coefficients of are equal and the constant terms are equal. This gives the following system of equations for , , and : H110022A H11001 2 B H11002 2 C H33527 H110021 3 A H11001 2 B H11002 C H33527 2 2 A H11001 B H11001 2 C H33527 1 CBA x x 2 2A H11001 B H11001 2Cx 2 x 2 H11001 2x H11002 1 H33527 H208492A H11001 B H11001 2CH20850x 2 H11001 H208493A H11001 2B H11002 CH20850x H11002 2A5 x 2 H11001 2x H11002 1 H33527 AH208492x H11002 1H20850H20849x H11001 2H20850 H11001 BxH20849x H11001 2H20850 H11001 CxH208492x H11002 1H208504 xH208492x H11002 1H20850H20849x H11001 2H20850 CBA x 2 H11001 2x H11002 1 xH208492x H11002 1H20850H20849x H11001 2H20850 H33527 A x H11001 B 2x H11002 1 H11001 C x H11001 2 3 2x 3 H11001 3x 2 H11002 2x H33527 xH208492x 2 H11001 3x H11002 2H20850 H33527 xH208492x H11002 1H20850H20849x H11001 2H20850 y x 2 H11001 2x H11002 1 2x 3 H11001 3x 2 H11002 2x dxV RH20849xH20850 QH20849xH20850 H33527 A 1 a 1 x H11001 b 1 H11001 A 2 a 2 x H11001 b 2 H11001H11080H11080H11080H11001 A k a k x H11001 b k 2 A 1 , A 2 , ..., A k QH20849xH20850 H33527 H20849a 1 x H11001 b 1 H20850H20849a 2 x H11001 b 2 H20850 H11080H11080H11080H20849a k x H11001 b k H20850 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS |||| 475 N Another method for finding , , and is given in the note after this example. CBA Solving, we get , , and , and so In integrating the middle term we have made the mental substitution , which gives and . M We can use an alternative method to find the coefficients , , and in Example 2. Equation 4 is an identity; it is true for every value of . Let’s choose values of that simplify the equation. If we put in Equation 4, then the second and third terms on the right side vanish and the equation then becomes , or . Likewise, gives and gives , so and . (You may object that Equation 3 is not valid for , , or , so why should Equation 4 be valid for those values? In fact, Equation 4 is true for all values of , even , , and . See Exercise 69 for the reason.) EXAMPLE 3 Find , where . SOLUTION The method of partial fractions gives and therefore Using the method of the preceding note, we put in this equation and get , so . If we put , we get , so . Thus Since , we can write the integral as See Exercises 55–56 for ways of using Formula 6. M CASE 11 N Q(x) is a product of linear factors, some of which are repeated. Suppose the first linear factor is repeated times; that is, occurs in the factorization of . Then instead of the single term in Equation 2, we A 1 H20862H20849a 1 x H11001 b 1 H20850QH20849xH20850 H20849a 1 x H11001 b 1 H20850 r rH20849a 1 x H11001 b 1 H20850 y dx x 2 H11002 a 2 H33527 1 2a ln H20895 x H11002 a x H11001 a H20895 H11001 C6 ln x H11002 ln y H33527 lnH20849xH20862yH20850 H33527 1 2a (ln H11341 x H11002 a H11341 H11002 ln H11341 x H11001 a H11341 ) H11001 C y dx x 2 H11002 a 2 H33527 1 2a y H20873 1 x H11002 a H11002 1 x H11001 a H20874 dx B H33527 H110021H20862H208492aH20850BH20849H110022aH20850 H33527 1x H33527 H11002aA H33527 1H20862H208492aH20850AH208492aH20850 H33527 1 x H33527 a AH20849x H11001 aH20850 H11001 BH20849x H11002 aH20850 H33527 1 1 x 2 H11002 a 2 H33527 1 H20849x H11002 aH20850H20849x H11001 aH20850 H33527 A x H11002 a H11001 B x H11001 a a HS33527 0 y dx x 2 H11002 a 2 H110022 1 2 x H33527 0x H110022 1 2 x H33527 0 C H33527 H11002 1 10 B H33527 1 5 10C H33527 H110021x H33527 H1100225BH208624 H33527 1 4 x H33527 1 2 A H33527 1 2 H110022A H33527 H110021 x H33527 0x x CBANOTE dx H33527 duH208622du H33527 2 dx u H33527 2x H11002 1 H33527 1 2 ln H11341 x H11341 H11001 1 10 ln H11341 2x H11002 1 H11341 H11002 1 10 ln H11341 x H11001 2 H11341 H11001 K y x 2 H11001 2x H11002 1 2x 3 H11001 3x 2 H11002 2x dx H33527 y H20873 1 2 1 x H11001 1 5 1 2x H11002 1 H11002 1 10 1 x H11001 2 H20874 dx C H33527 H11002 1 10 B H33527 1 5 A H33527 1 2 476 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION N We could check our work by taking the terms to a common denominator and adding them. N Figure 1 shows the graphs of the integrand in Example 2 and its indefinite integral (with ). Which is which?K H33527 0 FIGURE 1 _3 _2 2 3 would use By way of illustration, we could write but we prefer to work out in detail a simpler example. EXAMPLE 4 Find . SOLUTION The first step is to divide. The result of long division is The second step is to factor the denominator . Since , we know that is a factor and we obtain Since the linear factor occurs twice, the partial fraction decomposition is Multiplying by the least common denominator, , we get Now we equate coefficients: Solving, we obtain , , and , so M H33527 x 2 2 H11001 x H11002 2 x H11002 1 H11001 ln H20895 x H11002 1 x H11001 1 H20895 H11001 K H33527 x 2 2 H11001 x H11001 ln H11341 x H11002 1 H11341 H11002 2 x H11002 1 H11002 ln H11341 x H11001 1 H11341 H11001 K y x 4 H11002 2x 2 H11001 4x H11001 1 x 3 H11002 x 2 H11002 x H11001 1 dx H33527 y H20875 x H11001 1 H11001 1 x H11002 1 H11001 2 H20849x H11002 1H20850 2 H11002 1 x H11001 1 H20876 dx C H33527 H110021B H33527 2A H33527 1 H11002A H11001 B H11001 C H33527 0 A H11002 B H11002 2 C H33527 4 A H11001 B H11001 C H33527 0 H33527 H20849A H11001 CH20850x 2 H11001 H20849B H11002 2CH20850x H11001 H20849H11002A H11001 B H11001 CH20850 4 x H33527 AH20849x H11002 1H20850H20849x H11001 1H20850 H11001 BH20849x H11001 1H20850 H11001 CH20849x H11002 1H20850 2 8 H20849x H11002 1H20850 2 H20849x H11001 1H20850 4x H20849x H11002 1H20850 2 H20849x H11001 1H20850 H33527 A x H11002 1 H11001 B H20849x H11002 1H20850 2 H11001 C x H11001 1 x H11002 1 H33527 H20849x H11002 1H20850 2 H20849x H11001 1H20850 x 3 H11002 x 2 H11002 x H11001 1 H33527 H20849x H11002 1H20850H20849x 2 H11002 1H20850 H33527 H20849x H11002 1H20850H20849x H11002 1H20850H20849x H11001 1H20850 x H11002 1 QH208491H20850 H33527 0QH20849xH20850 H33527 x 3 H11002 x 2 H11002 x H11001 1 x 4 H11002 2x 2 H11001 4x H11001 1 x 3 H11002 x 2 H11002 x H11001 1 H33527 x H11001 1 H11001 4x x 3 H11002 x 2 H11002 x H11001 1 y x 4 H11002 2x 2 H11001 4x H11001 1 x 3 H11002 x 2 H11002 x H11001 1 dx x 3 H11002 x H11001 1 x 2 H20849x H11002 1H20850 3 H33527 A x H11001 B x 2 H11001 C x H11002 1 H11001 D H20849x H11002 1H20850 2 H11001 E H20849x H11002 1H20850 3 A 1 a 1 x H11001 b 1 H11001 A 2 H20849a 1 x H11001 b 1 H20850 2 H11001H11080H11080H11080H11001 A r H20849a 1 x H11001 b 1 H20850 r 7 SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS |||| 477 N Another method for finding the coefficients: Put in (8): . Put : . Put : .A H33527 B H11001 C H33527 1x H33527 0 C H33527 H110021x H33527 H110021 B H33527 2x H33527 1 CASE III N Q(x) contains irreducible quadratic factors, none of which is repeated. If has the factor , where , then, in addition to the partial fractions in Equations 2 and 7, the expression for will have a term of the form where and are constants to be determined. For instance, the function given by has a partial fraction decomposition of the form The term given in (9) can be integrated by completing the square and using the formula EXAMPLE 5 Evaluate . SOLUTION Since can’t be factored further, we write Multiplying by , we have Equating coefficients, we obtain Thus , , and and so In order to integrate the second term we split it into two parts: We make the substitution in the first of these integrals so that . We evaluate the second integral by means of Formula 10 with : M H33527 ln H11341 x H11341 H11001 1 2 lnH20849x 2 H11001 4H20850 H11002 1 2 tan H110021 H20849xH208622H20850 H11001 K y 2x 2 H11002 x H11001 4 xH20849x 2 H11001 4H20850 dx H33527 y 1 x dx H11001 y x x 2 H11001 4 dx H11002 y 1 x 2 H11001 4 dx a H33527 2 du H33527 2x dxu H33527 x 2 H11001 4 y x H11002 1 x 2 H11001 4 dx H33527 y x x 2 H11001 4 dx H11002 y 1 x 2 H11001 4 dx y 2x 2 H11002 x H11001 4 x 3 H11001 4x dx H33527 y H20873 1 x H11001 x H11002 1 x 2 H11001 4 H20874 dx C H33527 H110021B H33527 1A H33527 1 4A H33527 4C H33527 H110021A H11001 B H33527 2 H33527 H20849A H11001 BH20850x 2 H11001 Cx H11001 4A 2 x 2 H11002 x H11001 4 H33527 AH20849x 2 H11001 4H20850 H11001 H20849Bx H11001 CH20850x xH20849x 2 H11001 4H20850 2x 2 H11002 x H11001 4 xH20849x 2 H11001 4H20850 H33527 A x H11001 Bx H11001 C x 2 H11001 4 x 3 H11001 4x H33527 xH20849x 2 H11001 4H20850 y 2x 2 H11002 x H11001 4 x 3 H11001 4x dxV y dx x 2 H11001 a 2 H33527 1 a tan H110021 H20873 x a H20874 H11001 C10 x H20849x H11002 2H20850H20849x 2 H11001 1H20850H20849x 2 H11001 4H20850 H33527 A x H11002 2 H11001 Bx H11001 C x 2 H11001 1 H11001 Dx H11001 E x 2 H11001 4 f H20849xH20850 H33527 xH20862H20851H20849x H11002 2H20850H20849x 2 H11001 1H20850H20849x 2 H11001 4H20850H20852 BA Ax H11001 B ax 2 H11001 bx H11001 c 9 RH20849xH20850H20862QH20849xH20850 b 2 H11002 4ac H11021 0ax 2 H11001 bx H11001 cQH20849xH20850 478 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION EXAMPLE 6 Evaluate . SOLUTION Since the degree of the numerator is not less than the degree of the denominator, we first divide and obtain Notice that the quadratic is irreducible because its discriminant is . This means it can’t be factored, so we don’t need to use the partial fraction technique. To integrate the given function we complete the square in the denominator: This suggests that we make the substitution . Then, and ,so M Example 6 illustrates the general procedure for integrating a partial fraction of the form We complete the square in the denominator and then make a substitution that brings the integral into the form Then the first integral is a logarithm and the second is expressed in terms of . CASE IV N Q(x) contains a repeated irreducible quadratic factor. If has the factor , where , then instead of the single partial fraction (9), the sum A 1 x H11001 B 1 ax 2 H11001 bx H11001 c H11001 A 2 x H11001 B 2 H20849ax 2 H11001 bx H11001 cH20850 2 H11001H11080H11080H11080H11001 A r x H11001 B r H20849ax 2 H11001 bx H11001 cH20850 r 11 b 2 H11002 4ac H11021 0H20849ax 2 H11001 bx H11001 cH20850 r QH20849xH20850 tan H110021 y Cu H11001 D u 2 H11001 a 2 du H33527 C y u u 2 H11001 a 2 du H11001 D y 1 u 2 H11001 a 2 du where b 2 H11002 4ac H11021 0 Ax H11001 B ax 2 H11001 bx H11001 c NOTE H33527 x H11001 1 8 lnH208494x 2 H11002 4x H11001 3H20850 H11002 1 4 s2 tan H110021 H20873 2x H11002 1 s2 H20874 H11001 C H33527 x H11001 1 8 lnH20849u 2 H11001 2H20850 H11002 1 4 H11554 1 s2 tan H110021 H20873 u s2 H20874 H11001 C H33527 x H11001 1 4 y u u 2 H11001 2 du H11002 1 4 y 1 u 2 H11001 2 du H33527 x H11001 1 2 y 1 2 H20849u H11001 1H20850 H11002 1 u 2 H11001 2 du H33527 x H11001 1 4 y u H11002 1 u 2 H11001 2 du y 4x 2 H11002 3x H11001 2 4x 2 H11002 4x H11001 3 dx H33527 y H20873 1 H11001 x H11002 1 4x 2 H11002 4x H11001 3 H20874 dx x H33527 1 2 H20849u H11001 1H20850 du H33527 2 dxu H33527 2x H11002 1 4x 2 H11002 4x H11001 3 H33527 H208492x H11002 1H20850 2 H11001 2 b 2 H11002 4ac H33527 H1100232 H11021 0 4x 2 H11002 4x H11001 3 4x 2 H11002 3x H11001 2 4x 2 H11002 4x H11001 3 H33527 1 H11001 x H11002 1 4x 2 H11002 4x H11001 3 y 4x 2 H11002 3x H11001 2 4x 2 H11002 4x H11001 3 dx SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS |||| 479 occurs in the partial fraction decomposition of . Each of the terms in (11) can be integrated by first completing the square. EXAMPLE 7 Write out the form of the partial fraction decomposition of the function SOLUTION M EXAMPLE 8 Evaluate . SOLUTION The form of the partial fraction decomposition is Multiplying by , we have If we equate coefficients, we get the system which has the solution , , , , and . Thus M We note that sometimes partial fractions can be avoided when integrating a rational func- tion. For instance, although the integral y x 2 H11001 1 xH20849x 2 H11001 3H20850 dx H33527 ln H11341 x H11341 H11002 1 2 lnH20849x 2 H11001 1H20850 H11002 tan H110021 x H11002 1 2H20849x 2 H11001 1H20850 H11001 K H33527 y dx x H11002 y x x 2 H11001 1 dx H11002 y dx x 2 H11001 1 H11001 y x dx H20849x 2 H11001 1H20850 2 y 1 H11002 x H11001 2x 2 H11002 x 3 xH20849x 2 H11001 1H20850 2 dx H33527 y H20873 1 x H11002 x H11001 1 x 2 H11001 1 H11001 x H20849x 2 H11001 1H20850 2 H20874 dx E H33527 0D H33527 1C H33527 H110021B H33527 H110021A H33527 1 A H33527 1C H11001 E H33527 H1100212A H11001 B H11001 D H33527 2C H33527 H110021A H11001 B H33527 0 H33527 H20849A H11001 BH20850x 4 H11001 Cx 3 H11001 H208492A H11001 B H11001 DH20850x 2 H11001 H20849C H11001 EH20850x H11001 A H33527 AH20849x 4 H11001 2x 2 H11001 1H20850 H11001 BH20849x 4 H11001 x 2 H20850 H11001 CH20849x 3 H11001 xH20850 H11001 Dx 2 H11001 Ex H11002x 3 H11001 2x 2 H11002 x H11001 1 H33527 AH20849x 2 H11001 1H20850 2 H11001 H20849Bx H11001 CH20850xH20849x 2 H11001 1H20850 H11001 H20849Dx H11001 EH20850x xH20849x 2 H11001 1H20850 2 1 H11002 x H11001 2x 2 H11002 x 3 xH20849x 2 H11001 1H20850 2 H33527 A x H11001 Bx H11001 C x 2 H11001 1 H11001 Dx H11001 E H20849x 2 H11001 1H20850 2 y 1 H11002 x H11001 2x 2 H11002 x 3 xH20849x 2 H11001 1H20850 2 dx H33527 A x H11001 B x H11002 1 H11001 Cx H11001 D x 2 H11001 x H11001 1 H11001 Ex H11001 F x 2 H11001 1 H11001 Gx H11001 H H20849x 2 H11001 1H20850 2 H11001 Ix H11001 J H20849x 2 H11001 1H20850 3 x 3 H11001 x 2 H11001 1 xH20849x H11002 1H20850H20849x 2 H11001 x H11001 1H20850H20849x 2 H11001 1H20850 3 x 3 H11001 x 2 H11001 1 xH20849x H11002 1H20850H20849x 2 H11001 x H11001 1H20850H20849x 2 H11001 1H20850 3 RH20849xH20850H20862QH20849xH20850 480 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION N It would be extremely tedious to work out by hand the numerical values of the coefficients in Example 7. Most computer algebra systems, however, can find the numerical values very quickly. For instance, the Maple command or the Mathematica command gives the following values: I H33527 H11002 1 2 , J H33527 1 2 E H33527 15 8 , F H33527 H11002 1 8 , G H33527 H H33527 3 4 , A H33527 H110021, B H33527 1 8 , C H33527 D H33527 H110021, Apart[f] convertH20849f, parfrac, xH20850 N In the second and fourth terms we made the mental substitution .u H33527 x 2 H11001 1 could be evaluated by the method of Case III, it’s much easier to observe that if , then and so RATIONALIZING SUBSTITUTIONS Some nonrational functions can be changed into rational functions by means of appropri- ate substitutions. In particular, when an integrand contains an expression of the form , then the substitution may be effective. Other instances appear in the exercises. EXAMPLE 9 Evaluate . SOLUTION Let . Then , so and . Therefore We can evaluate this integral either by factoring as and using partial fractions or by using Formula 6 with : M H33527 2sx H11001 4 H11001 2ln H20895 sx H11001 4 H11002 2 sx H11001 4 H11001 2 H20895 H11001 C H33527 2u H11001 8 H11554 1 2 H11554 2 ln H20895 u H11002 2 u H11001 2 H20895 H11001 C y sx H11001 4 x dx H33527 2 y du H11001 8 y du u 2 H11002 4 a H33527 2 H20849u H11002 2H20850H20849u H11001 2H20850u 2 H11002 4 H33527 2 y H20873 1 H11001 4 u 2 H11002 4 H20874 du y sx H11001 4 x dx H33527 y u u 2 H11002 4 2u du H33527 2 y u 2 u 2 H11002 4 du dx H33527 2u dux H33527 u 2 H11002 4u 2 H33527 x H11001 4u H33527sx H11001 4 y sx H11001 4 x dx u H33527s n tH20849xH20850s n tH20849xH20850 y x 2 H11001 1 xH20849x 2 H11001 3H20850 dx H33527 1 3 ln H11341 x 3 H11001 3x H11341 H11001 C du H33527 H208493x 2 H11001 3H20850 dxu H33527 xH20849x 2 H11001 3H20850 H33527 x 3 H11001 3x SECTION 7.4 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS |||| 481 (a) (b) 6. (a) (b) 7–38 Evaluate the integral. 7. 8. 9. 10. y 1 H20849t H11001 4H20850H20849t H11002 1H20850 dt y x H11002 9 H20849x H11001 5H20850H20849x H11002 2H20850 dx y r 2 r H11001 4 dr y x x H11002 6 dx 1 x 6 H11002 x 3 x 4 H20849x 3 H11001 xH20850H20849x 2 H11002 x H11001 3H20850 t 4 H11001 t 2 H11001 1 H20849t 2 H11001 1H20850H20849t 2 H11001 4H20850 2 x 4 x 4 H11002 1 5. 1–6 Write out the form of the partial fraction decomposition of the function (as in Example 7). Do not determine the numerical values of the coefficients. 1. (a) (b) 2. (a) (b) 3. (a) (b) 4. (a) (b) 2x H11001 1 H20849x H11001 1H20850 3 H20849x 2 H11001 4H20850 2 x 3 x 2 H11001 4x H11001 3 1 H20849x 2 H11002 9H20850 2 x 4 H11001 1 x 5 H11001 4x 3 x 2 x 2 H11001 x H11001 2 x x 2 H11001 x H11002 2 1 x 3 H11001 2x 2 H11001 x 2x H20849x H11001 3H20850H208493x H11001 1H20850 EXERCISES7.4 482 |||| CHAPTER 7 TECHNIQUES OF INTEGRATION 48. 49. 50. 51–52 Use integration by parts, together with the techniques of this section, to evaluate the integral. 51. 52. ; 53. Use a graph of to decide whether is positive or negative. Use the graph to give a rough estimate of the value of the integral and then use partial fractions to find the exact value. ; 54. Graph both and an antiderivative on the same screen. 55–56 Evaluate the integral by completing the square and using Formula 6. 56. 57. The German mathematician Karl Weierstrass (1815–1897) noticed that the substitution will convert any rational function of and into an ordinary rational function of . (a) If , , sketch a right triangle or use trigonometric identities to show that (b) Show that (c) Show that 58–61 Use the substitution in Exercise 57 to transform the inte- grand into a rational function of and then evaluate the integral. 58. 59. 60. y H9266H208622 H9266H208623 1 1 H11001 sin x H11002 cos x dx y 1 3 sin x H11002 4 cos x dx y dx 3 H11002 5 sin x t dx H33527 2 1 H11001 t 2 dt cos x H33527 1 H11002 t 2 1 H11001 t 2 and sin x H33527 2t 1 H11001 t 2 cos H20873 x 2 H20874 H33527 1 s1 H11001 t 2 and sin H20873 x 2 H20874 H33527 t s1 H11001 t 2 H11002H9266 H11021 x H11021 H9266t H33527 tanH20849xH208622H20850 t cos xsin x t H33527 tanH20849xH208622H20850 y 2x H11001 1 4x 2 H11001 12x H11002 7 dx y dx x 2 H11002 2x 55. y H33527 1H20862H20849x 3 H11002 2x 2 H20850 x 2 0 f H20849xH20850 dx f H20849xH20850 H33527 1H20862H20849x 2 H11002 2x H11002 3H20850 y x tan H110021 x dx y lnH20849x 2 H11002 x H11001 2H20850 dx y e x H20849e x H11002 2H20850H20849e 2x H11001 1H20850 dx y sec 2 t tan 2 t H11001 3 tan t H11001 2 dt y cos x sin 2 x H11001 sin x dx y e 2x e 2x H11001 3e x H11001 2 dx47. 12. 13. 14. 15. 16. 18. 19. 20. 21. 22. 23. 24. 26. 27. 28. 30. 32. 33. 34. 35. 36. 37. 38. 39–50 Make a substitution to express the integrand as a rational function and then evaluate the integral. 39. 40. 41. 42. 44. 45. [Hint: Substitute .] 46. y s1 H11001 sx x dx u H33527 6 sx y 1 sx H11002 s 3 x dx y 3 1H208623 sx x 2 H11001 x dx y x 3 s 3 x 2 H11001 1 dx43. y 1 0 1 1 H11001 s 3 x dx y 16 9 sx x H11002 4 dx y dx 2sx H11001 3 H11001 x y 1 xsx H11001 1 dx y x 3 H11001 2x 2 H11001 3x H11002 2 H20849x 2 H11001 2x H11001 2H20850 2 dx y x 2 H11002 3x H11001 7 H20849x 2 H11002 4x H11001 6H20850 2 dx y x 4 H11001 3x 2 H11001 1 x 5 H11001 5x 3 H11001 5x dx y dx xH20849x 2 H11001 4H20850 2 y x 3 x 3 H11001 1 dx y 1 0 x 3 H11001 2x x 4 H11001 4x 2 H11001 3 dx y 1 0 x x 2 H11001 4x H11001 13 dx y 1 x 3 H11002 1 dx31. y 3x 2 H11001 x H11001 4 x 4 H11001 3x 2 H11001 2 dx y x H11001 4 x 2 H11001 2x H11001 5 dx29. y x 2 H11002 2x H11002 1 H20849x H11002 1H20850 2 H20849x 2 H11001 1H20850 dx y x 3 H11001 x 2 H11001 2x H11001 1 H20849x 2 H11001 1H20850H20849x 2 H11001 2H20850 dx y x 2 H11001 x H11001 1 H20849x 2 H11001 1H20850 2 dx y 10 H20849x H11002 1H20850H20849x 2 H11001 9H20850 dx25. y x 2 H11002 x H11001 6 x 3 H11001 3x dx y 5x 2 H11001 3x H11002 2 x 3 H11001 2x 2 dx y ds s 2 H20849s H11002 1H20850 2 y x 3 H11001 4 x 2 H11001 4 dx y x 2 H11002 5x H11001 16 H208492x H11001 1H20850H20849x H11002 2H20850 2 dx y 1 H20849x H11001 5H20850 2 H20849x H11002 1H20850 dx y x 2 H11001 2x H11002 1 x 3 H11002 x dx y 2 1 4y 2 H11002 7y H11002 12 yH20849y H11001 2H20850H20849y H11002 3H20850 dy17. y 1 0 x 3 H11002 4x H11002 10 x 2 H11002 x H11002 6 dx y 4 3 x 3 H11002 2x 2 H11002 4 x 3 H11002 2x 2 dx y 1 H20849x H11001 aH20850H20849x H11001 bH20850 dx y ax x 2 H11002 bx dx y 1 0 x H11002 1 x 2 H11001 3x H11001 2 dx y 3 2 1 x 2 H11002 1 dx11. SECTION 7.5 STRATEGY FOR INTEGRATION |||| 483 67. (a) Use a computer algebra system to find the partial fraction decomposition of the function (b) Use part (a) to find (by hand) and compare with the result of using the CAS to integrate directly. Com- ment on any discrepancy. 68. (a) Find the partial fraction decomposition of the function (b) Use part (a) to find and graph and its indefinite integral on the same screen. (c) Use the graph of to discover the main features of the graph of . 69. Suppose that , and are polynomials and for all except when . Prove that for all . [Hint: Use continuity.] 70. If is a quadratic function such that and is a rational function, find the value of .f H11032H208490H20850 y f H20849xH20850 x 2 H20849x H11001 1H20850 3 dx f H208490H20850 H33527 1f x FH20849xH20850 H33527 GH20849xH20850QH20849xH20850 H33527 0x FH20849xH20850 QH20849xH20850 H33527 GH20849xH20850 QH20849xH20850 QF, G x f H20849xH20850 dx f fx f H20849xH20850 dx f H20849xH20850 H33527 12x 5 H11002 7x 3 H11002 13x 2 H11001 8 100x 6 H11002 80x 5 H11001 116x 4 H11002 80x 3 H11001 41x 2 H11002 20x H11001 4 CAS f x f H20849xH20850 dx f H20849xH20850 H33527 4x 3 H11002 27x 2 H11001 5x H11002 32 30x 5 H11002 13x 4 H11001 50x 3 H11002 286x 2 H11002 299x H11002 70 CAS 61. 62–63 Find the area of the region under the given curve from 1 to 2. 62. 63. 64. Find the volume of the resulting solid if the region under the curve from to is rotated about (a) the -axis and (b) the -axis. 65. One method of slowing the growth of an insect population without using pesticides is to introduce into the population a number of sterile males that mate with fertile females but produce no offspring. If represents the number of female insects in a population, the number of sterile males introduced each generation, and the population’s natural growth rate, then the female population is related to time by Suppose an insect population with 10,000 females grows at a rate of and 900 sterile males are added. Evaluate the integral to give an equation relating the female population to time. (Note that the resulting equation can’t be solved explic- itly for .) 66. Factor as a difference of squares by first adding and subtracting the same quantity. Use this factorization to evalu- ate .x 1H20862H20849x 4 H11001 1H20850 dx x 4 H11001 1 P r H33527 0.10 t H33527 y P H11001 S PH20851H20849r H11002 1H20850P H11002 SH20852 dP t r S P yx x H33527 1x H33527 0y H33527 1H20862H20849x 2 H11001 3x H11001 2H20850 y H33527 x 2 H11001 1 3x H11002 x 2 y H33527 1 x 3 H11001 x y H9266H208622 0 sin 2x 2 H11001 cos x dx STRATEGY FOR INTEGRATION As we have seen, integration is more challenging than differentiation. In finding the deriv- ative of a function it is obvious which differentiation formula we should apply. But it may not be obvious which technique we should use to integrate a given function. Until now individual techniques have been applied in each section. For instance, we usually used substitution in Exercises 5.5, integration by parts in Exercises 7.1, and partial fractions in Exercises 7.4. But in this section we present a collection of miscellaneous inte- grals in random order and the main challenge is to recognize which technique or formula to use. No hard and fast rules can be given as to which method applies in a given situation, but we give some advice on strategy that you may find useful. A prerequisite for strategy selection is a knowledge of the basic integration formulas. In the following table we have collected the integrals from our previous list together with several additional formulas that we have learned in this chapter. Most of them should be memorized. It is useful to know them all, but the ones marked with an asterisk need not be 7.5 James Stewart Stewart - Calculus - Early Transcedentals 6e calculus

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