newmaster5 - 5 5.1 Techniques of Anti-differentiation Differential Notation 1 Calculate the differential of the following functions by using the

newmaster5 - 5 5.1 Techniques of Anti-differentiation...

This preview shows page 1 - 3 out of 16 pages.

5 Techniques of Anti-differentiation 5.1 Differential Notation 1 Calculate the differential of the following functions by using the definition dy = y 0 ( x ) dx. Express the result in terms of the product between y 0 ( x ) and the differential of x , dx . For example, given y ( x ) = 3 x + sin(2 x ) , its differential is dy = y 0 ( x ) dx = (3 + 2 cos(2 x )) dx. (a) f ( x ) = e x 2 (b) f ( x ) = ( x + 1) 2 (c) f ( x ) = x (d) f ( x ) = arcsin( x ) (e) f ( x ) = x 2 + 3 x + 1 (f) f ( x ) = cos(2 x ) (g) y = x 6 + 2 x 4 - 2 x (h) y = ( x - 2) 2 ( x + 1) 5 (i) y = x/ ( x + 3) Solution (a) d ( e x 2 ) = 2 xe x 2 dx (b) d (( x + 1) 2 ) = 2( x + 1) dx (c) d ( x ) = (1 / 2) x - 1 / 2 dx (d) d (arcsin( x )) = 1 1 - x 2 dx (e) d ( x 2 + 3 x + 1) = (2 x + 3) dx (f) d (cos(2 x )) = - 2 sin(2 x ) dx (g) dy = (6 x 5 + 8 x 3 - 2) dx (h) dy = (2( x - 2)( x + 1) 5 + 5( x - 2) 2 ( x + 1) 4 ) dx (i) dy = ((1( x + 3) - x ) / ( x + 3) 2 ) dx = (3 / ( x + 3) 2 ) dx 5.2 Differential Notation 2 Given the differential of a function y in terms of the product between its derivative y 0 ( x ) and the differential of x , ( dx ), express the same differential in terms of the the differential of y itself, i.e., dy . Then, use the result combined with the fundamental theorem of calculus to calculate the corresponding indefinite integrals. For example, if we know that the differential of a function y ( x ) is given by (3 + 2 sin(2 x )) dx =? , (Or equivalently , Z (3 + 2 sin(2 x )) dx =?) we now need to figure out that y ( x ) = 3 x - cos(2 x ) and express the differential given above in terms of the differential of y itself (3 + 2 sin(2 x )) dx = d (3 x - cos(2 x )) . 1
Using this result and the fundamental theorem of calculus, we can solve the following Z (3 + 2 sin(2 x )) dx = Z d (3 x - cos(2 x )) = 3 x - cos(2 x ) + C. (Those marked by a star are a little bit more difficult since they involve composite functions.) (a) x 3 dx =? , R x 3 dx =? (b) x 3 / 2 dx =? , R x 3 / 2 dx =? (c) 1 x 3 dx =? , R 1 x 3 dx =? (d) xdx =? , R xdx =? (e) 1 x dx =? , R 1 x dx =? (f) e - 2 x dx =? , R e - 2 x dx =? (g) ( x + 1) 2 dx =? , R ( x + 1) 2 dx =? (h) 1 x +3) 2 dx =? , R 1 x +3) 2 dx =? (i) cos(2x)dx=?,Rcos(2x)dx=?(j) sec2(x)dx=?,Rsec2(x)dx=? (i) cos(2 x ) dx = d sin(2 x ) 2 , R cos(2 x ) dx = sin(2 x ) 2 + C. (j) sec 2 ( x ) dx = d tan( x ) , R sec 2 ( x ) dx = tan( x ) + C (k) 1 x dx = d ln x, R 1 x dx = ln x + C. (l) 1 1+ x 2 dx = d tan - 1 x, R 1 1+ x 2 dx = tan - 1 x + C. (m) 2 xe x 2 dx = de x 2 , R 2 xe x 2 dx = R de x 2 = e x 2 + C. (n) 3 x 2 cos( x 3 ) dx = d sin( x 3 ), R 3 x 2 cos( x 3 ) dx = R d sin( x 3 ) + sin( x 3 ) + C. (o) 2 x 1+ x 2 dx = d ln(1 + x 2 ), R 2 x 1+ x 2 dx = R d ln(1 + x 2 ) = ln(1 + x 2 ) + C. (p) 2 x 1+ x 4 dx = dx 2 1+( x 2 ) 2 = d tan - 1 ( x 2 ), R 2 x 1+ x 4 dx = R d tan - 1 ( x 2 ) = tan - 1 ( x 2 ) + C. 5.3 Substitution 1 (a) Evaluate the following indefinite integrals using the substitution method: Z sin(3 x ) dx, Z cos( x ) x dx, Z x 3 x 4 + 1 dx, Z 4 1 + 2 x dx.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture