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N=n => width = 1/n US = =1/n 3 [1+4+9+….+n 2 ] US =

. N=n => width = 1/n US = =1/n 3 [1+4+9+….+n 2 ] US =. =. Definite Integral. We will define to be the limit as n approaches oo of where D x i = (b-a)/n and is any point in the ith interval. where D x i = (b-a)/n

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N=n => width = 1/n US = =1/n 3 [1+4+9+….+n 2 ] US =

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  1. . N=n => width = 1/n US = =1/n3[1+4+9+….+n2] US =

  2. =

  3. DefiniteIntegral We will define to be the limit as n approaches oo of where Dxi = (b-a)/n and is any point in the ith interval.

  4. where Dxi = (b-a)/n and is any point in the ith interval, [xi-1,xi].

  5. > 0 when f(x) > 0, but it is negative when f(x)<0. We will define to be the limit as n approaches oo of and is any point in the ith interval, [xi-1,xi].

  6. DefinitionTheorems 1. 2. 3. 4.

  7. DefinitionTheorems 5. 6. 7. 8.

  8. If f(x) >= 0 on [a,b] then is the area under f(x) and over the x-axis between a and b.

  9. If f(x) <= 0 on [a,b] then is the negative of the area over f(x) and under the x-axis between a and b.

  10. [ • 0.50 • 0.1

  11. [ • 2.0 • 0.1

  12. ] • 0.0 • 0.1

  13. [ • 1.0 • 0.1

  14. [ • 1.5 • 0.1

  15. 0

  16. ] • 1.0 • 0.1

  17. where Dxi = (b-a)/n and is any point in the ith interval, [xi-1,xi]. If the interval is [-4, 4] evaluate 8p

  18. Pi = 3.14 • 6.28 • 0.1

  19. Pi = 3.14 • -6.28 • 0.1

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