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anonymous

  • one year ago

Intergration problem

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    i have prblem with the "e"

  3. anonymous
    • one year ago
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    if it was only x it would be easyier

  4. geerky42
    • one year ago
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    You know what \(\dfrac{\mathrm d}{\mathrm dx}~e^{2x}\) is, right?

  5. anonymous
    • one year ago
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    yes

  6. geerky42
    • one year ago
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    To find integral of \(5e^{2x}\), you want to look for value \(c\) in \(c~e^{2x}\) such that taking derivaitve it would give you \(5e^{2x}\)

  7. geerky42
    • one year ago
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    Since \(\dfrac{\mathrm d}{\mathrm dx}~e^{2x} = 2e^{2x}\), you have \(\dfrac{\mathrm d}{\mathrm dx}~c~e^{2x} = 2c~e^{2x}\). Here, we have \(2c = 5\)

  8. anonymous
    • one year ago
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    i thought we have to take the antider wouldnt it be 5/2e^2x?

  9. geerky42
    • one year ago
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    So that c is 5/2 Therefore integral of \(5e^{2x}\) is \(\dfrac{5e^{2x}}{2}\)

  10. anonymous
    • one year ago
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    yes what do i do after ? for substition

  11. geerky42
    • one year ago
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    substitution? Well, you basically have \(\left.\dfrac{5}{2}e^{2x}~\right|_1^8 = \left(\dfrac{5}{2}e^{2(8)}\right)-\left(\dfrac{5}{2}e^{2(1)}\right)\)

  12. anonymous
    • one year ago
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    ok

  13. geerky42
    • one year ago
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    You know how to evaluate \(\displaystyle \int_1^8 5e^{2x}~\mathrm dx?\), right? do same for other terms.

  14. anonymous
    • one year ago
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    for 4/x i get 0 ?

  15. anonymous
    • one year ago
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    i am right ?

  16. anonymous
    • one year ago
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    -1+1 =0

  17. geerky42
    • one year ago
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    \[\int\dfrac{4}{x}\mathrm~dx = 4\ln(8) - 4\ln(1)\]

  18. anonymous
    • one year ago
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    idk i might be right , but my final solution is s/2e^14 + 196607.25

  19. anonymous
    • one year ago
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    damn i am wrong

  20. geerky42
    • one year ago
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    196607.25 isn't exact value?

  21. anonymous
    • one year ago
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    k after simplity everything i get 5/2e^2x - 4lnx-3/4x^3/4(x)

  22. anonymous
    • one year ago
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    k after simplity everything i get 5/2e^2x - 4lnx+3/4x^3/4(x

  23. anonymous
    • one year ago
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    @geerky42

  24. anonymous
    • one year ago
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    @ikram002p

  25. anonymous
    • one year ago
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    @ganeshie8

  26. anonymous
    • one year ago
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    @satellite73

  27. anonymous
    • one year ago
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    @campbell_st

  28. anonymous
    • one year ago
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    @satellite73

  29. anonymous
    • one year ago
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    @triciaal

  30. anonymous
    • one year ago
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    \[\int\limits \left(\sqrt[3]{x}+5 e^{2 x}-\frac{4}{x}\right) \, dx=\frac{3 x^{4/3}}{4}+\frac{5 e^{2 x}}{2}-4 \log (x) \]\[\frac{\partial \left(\frac{3 x^{4/3}}{4}+\frac{5 e^{2 x}}{2}-4 \log (x)\right)}{\partial x}=\sqrt[3]{x}+5 e^{2 x}-\frac{4}{x} \]

  31. anonymous
    • one year ago
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    Sounds good

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