Change of Variables: Indefinite Integrals

Transcription

1 Change of Variables: Indefinite Integrals Mathematics 11: Lecture 39 Dan Sloughter Furman University November 29, 2007 Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

2 : reversing the chain rule Suppose we wish to evaluate 2x 1 + x 2 dx. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

3 : reversing the chain rule Suppose we wish to evaluate 2x 1 + x 2 dx. Since 2x is the derivative of 1 + x 2 and xdx = 2 3 x c, we might guess 2x 1 + x 2 dx = 2 3 (1 + x 2 ) c, which we may verify by differentiation. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

4 Note: if we wanted to evaluate x 1 + x 2 dx we might still guess 2 3 (1 + x 2 ) 3 2. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

5 Note: if we wanted to evaluate x 1 + x 2 dx we might still guess However, 2 3 (1 + x 2 ) 3 2. ( ) d 2 dx 3 (1 + x 2 ) 3 2 = 2x 1 + x 2. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

6 (cont d) We could correct our guess by multiplying by 1 2. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

7 (cont d) We could correct our guess by multiplying by 1 2. That is, x 1 + x 2 dx = 1 3 (1 + x 2 ) c, which we may check by differentiation. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

8 Substitution Note: the first example above is of the form f (g(x))g (x)dx, where f (x) = x and g(x) = 1 + x 2. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

9 Substitution Note: the first example above is of the form f (g(x))g (x)dx, where f (x) = x and g(x) = 1 + x 2. In general, if F is an antiderivative of f, then, by the chain rule, f (g(x))g (x)dx = F (g(x)) + c. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

10 Leibniz notation Using Leibniz notation, if u = g(x), then f (u) du dx = F (u) + c. dx Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

11 Leibniz notation Using Leibniz notation, if u = g(x), then f (u) du dx = F (u) + c. dx But f (u)du = F (u) + c, so we have f (g(x)) g (x)dx = }{{}}{{} u du f (u) du dx dx = f (u)du. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

12 Leibniz notation Using Leibniz notation, if u = g(x), then f (u) du dx = F (u) + c. dx But f (u)du = F (u) + c, so we have f (g(x)) g (x)dx = }{{}}{{} u du f (u) du dx dx = f (u)du. Note: symbolically (and only symbolically), we may think of substituting u for g(x) and du for g (x)dx. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

13 To evaluate 2x 1 + x 2 dx, we make the substitution u = 1 + x 2, from which we have du dx = 2x or du = 2xdx. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

14 To evaluate 2x 1 + x 2 dx, we make the substitution u = 1 + x 2, from which we have du dx = 2x or du = 2xdx. Hence 2x 1 + x 2 dx = udu = 2 3 u c = 2 3 (1 + x 2 ) c. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

15 To evaluate x 1 + x 2 dx, we make the substitution u = 1 + x 2 du = 2xdx, from which it follows that 1 du = xdx. 2 Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

16 To evaluate x 1 + x 2 dx, we make the substitution u = 1 + x 2 du = 2xdx, from which it follows that 1 du = xdx. 2 Then we have x 1 + x 2 dx = 1 udu 1 = 2 3 u c = 3 (1 + x 2 ) c, as we saw above. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

17 To evaluate x 2 sin(4x 3 )dx, we make the substitution u = 4x 3 du = 12x 2 dx, or 1 12 du = x 2 dx. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

18 To evaluate x 2 sin(4x 3 )dx, we make the substitution u = 4x 3 du = 12x 2 dx, or 1 12 du = x 2 dx. Then x 2 sin(4x 3 )dx = 1 12 sin(u)du = 1 12 cos(u) + c = 1 12 cos(4x 3 ) + c. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

19 To evaluate sin 2 (x) cos(x)dx, we make the substitution u = sin(x) du = cos(x)dx. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

20 To evaluate sin 2 (x) cos(x)dx, we make the substitution u = sin(x) du = cos(x)dx. Then sin 2 (x) cos(x)dx = u 2 du = 1 3 u3 + c = 1 3 sin3 (x) + c. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

21 4x To evaluate + 5dx, we make the substitution u = 4x + 5 du = 4dx, or 1 du = dx. 4 Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

22 4x To evaluate + 5dx, we make the substitution u = 4x + 5 du = 4dx, or 1 du = dx. 4 Then 4x 1 udu 2 + 5dx = = 4 12 u c = 6 (4x + 5) c. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

23 To evaluate x 1 + xdx, we make the substitution u = 1 + x du = dx. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

24 To evaluate x 1 + xdx, we make the substitution Then, using x = u 1, x 1 + xdx = u = 1 + x du = dx. (u 1) udu = = 2 5 u u c = 2 5 (1 + x) (1 + x) c. u 3 2 du udu Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13

25 (cont d) Note: substitution worked in this problem, although it did not involve reversing the chain rule. Dan Sloughter (Furman University) Change of Variables: Indefinite Integrals November 29, / 13