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anonymous

  • 5 years ago

calculate the integral of the following using:

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  1. anonymous
    • 5 years ago
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    the standard form \[\int\limits_{}^{}f'(x)e ^{f(x)}dx=e ^{f(x)}+c , or \int\limits_{}^{}e ^{u}(du/dx)dx = e ^{u}+c\]

  2. anonymous
    • 5 years ago
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    \[(a) 2xe ^{x ^{2}+3}\] \[(b)(10x-2)e ^{5x ^{2}-2x}\] \[(c)(3x+2)e ^{3x ^{2}+4x+1}\] \[(d)(x ^{2}-2x)e ^{x ^{3}-3x ^{2}}\]

  3. watchmath
    • 5 years ago
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    What have you try with these problems?

  4. anonymous
    • 5 years ago
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    yea i have i can do (a),(b) but c and d i got the wrong answers

  5. watchmath
    • 5 years ago
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    You need to adjust a little bit for c) and d). For example in c) if you use \(u=3x^2+4x\) then \(du=6x+4=2(3x+2)\, dx\). So your integral become \(\frac{1}{2}\int e^u\, du\)

  6. anonymous
    • 5 years ago
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    oh i understand thanks but what happens to the (3x+2) inside the bracket

  7. watchmath
    • 5 years ago
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    \((3x+2)\,dx\) become \(\frac{1}{2}du\)

  8. anonymous
    • 5 years ago
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    ok so how about this one \[(6x ^{2}-8x+6)e ^{x ^{3}-2x ^{2}+3x-5}\]

  9. watchmath
    • 5 years ago
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    what is the derivative of \(x^3-2x^2+3x-5\) ?

  10. anonymous
    • 5 years ago
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    3x^2-4x+3

  11. watchmath
    • 5 years ago
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    Good! How that derivative relates to \(6x^2-8x+6\) ?

  12. anonymous
    • 5 years ago
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    its half of it right

  13. anonymous
    • 5 years ago
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    the coefficients are doubled

  14. watchmath
    • 5 years ago
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    Yes great! Then for \(f(x)=x^3-2x^2+3x-5\) the integral is \(\frac{1}{2}\int f'(x)e^{f(x)}\,dx\)

  15. watchmath
    • 5 years ago
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    Sorry your were right :)

  16. watchmath
    • 5 years ago
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    So \(2\int f'(x)e^{f(x)}\,dx\).

  17. anonymous
    • 5 years ago
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    oh i see how it works out now lol:) ok thanks ill have shot at the others thankyou

  18. watchmath
    • 5 years ago
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    Great! :D

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