Substitution & Definite integrals

Transcript

1. MATH 122 Substitution and the Definite Integral On this worksheet you will use substitution, as well as the other integration rules, to evaluate the the given definite and indefinite integrals. Steps for integration by Substitution Steps for finding the Definite Integral 1. Determine u: think parentheses and denominators 1. Using substitution or otherwise, find an antiderivative F (x) 2. Find du dx 2. Using the given limits of integration, find F (b) − F (a). Remember: b is the upper limit and a is the lower limit. 3. Rearrange du dx until you can make a substitution 4. Make the substitution to obtain an integral in u Be careful to evaluate −F (a) correctly (distribute the negative accordingly) 5. Integrate with respect to u Your answer should be a number 6. Substitute u back to be left with an expression in terms of x If you make a substitution, remember to substitute back before plugging in your limits of integration Example 1: Z
   Find 4x x2 + 1 )5 dx. Observe that if u = x2 + 1 then Z =⇒ Z 4x x2 + 1 So
   5 dx = du = 2x and so dx du = 2x dx =⇒ 2 du = 4x dx Z
   5 1 1 2 4x x + 1 dx = 2u5 du = u6 + C = (x2 + 1)6 + C. 3 3 1 2 (x + 1)6 + C. 3 Example 2: Z 5 Find 9x2 + 10x + 3 dx. 1 Z 5 9x2 + 10x + 3 dx = 3x3 + 5x2 + 3x   = 3(5)3 + 5(5)2 + 3(5) − 3(1)3 + 5(1)2 + 3(1) = 515 − 11 = 504 1 1 Z 5 5 9x2 + 10x + 3 dx = 504. So 1 Example 3: Z 2 Find 0 √ 2x dx. x2 + 4 Observe that if u = x2 + 4 then du = 2x and so dx du = 2x dx Z 2 =⇒ √ 0 Z 2 So 0 2x dx = x2 + 4 √ Z x=2 u−1/2 du = 2u1/2 x=0 √  2x dx = 4 2−1 . x2 + 4 x=2 x=0 = 2 x2 + 4
   1/2 2 0 = 2 22 + 4
   1/2 − 2 02 + 4
   1/2 √  √ = 2 8−4 = 4 2−1 .
2. MATH 122 Z 2 1. 1 Z 2. Z 7 1 dr r2 12s + 1 ds (2x − 1)6 dx 18. z dz (2 − z)(2 + z) 19. 10. 8 11. 4x 2x2 + 1 6. 12.
   5 dx 3 t dt 2ex 13. 3s2 + 2 ds 3 2s + 4s + 3 6x 3x2 + 5 Z 14. 3y 3x2 dx 3 x +8 15. dx Z 4 p 2 x3 6x e 1 √ 2 x (1 + 1 4 − y 2 dy Z 2 x dx 2x2 + 3 Z 1 16. dx 3 dx * x ln(x) 21. Z 2 Z ex + 1 2 0 Z 1 8. Z 0 4tet dt Z 9 0 e3x dx 4 − e3x 20. Z 3 Z Z − ln(2) Z 6y dy 7. 6t dt 2t2 + 5 0 Z 0 5. Z −2 Z 6 Z Page 2 of 2 17. Z 1 (4x + 1) dx 4. 2 9. 3 dx 3x + 5 Z 3. Substitution and the Definite Integral 22. √ 2 dx * x) x(x − 1)4 dx * 1 Z 23. dx 0 2x (4x + 5) 3 dx * Answers 1. Ans. 0.5 9. Ans. 1413 17. u = 2t2 + 5, Ans. 32 ln(2t2 + 5) + C 2. u = 3x + 5, Ans. ln |3x + 5| + C 10. u = 2x − 1, Ans. 71 1 3. u = 4x + 1)8 , Ans. 36 (4x + 1)9 11. u = 4 − z 2 , Ans. − 12 ln |4 − z 2 | + C 18. u = 4−e3x , Ans. − 31 ln |4−e3x |+C
   19. u = ex + 1, Ans. 2 ln 43 4. Ans. 108 12. u = t2 , Ans. 2et + C 5. u = 2x2 + 1, Ans. 16 (2x2 + 1)6 + C 13. u = 3x2 + 5, Ans. ln(3x2 + 5) + C 2 2 6. Ans. −1620 14. u = 4 − y , Ans. 8 7. u = 2s3 + 4s + 3, Ans. ln(3) 2 15. u = 2x2 +3, Ans. 41 ln(2x2 +3)+C 8. u = x3 + 8, Ans. ln |x3 + 8| + C 16. u = x3 , Ans. 2e − 2 20. u = ln(x), Ans. 3 ln | ln |x|| + C 21. u = 1 + √ x, Ans. 16 22. u = x − 1, Ans. 11 30 8x+5 23. u = 4x + 5, Ans. − 16(4x+5) 2 + C The End.