anonymous
  • anonymous
calculate the integral of the following using:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
the standard form \[\int\limits_{}^{}f'(x)e ^{f(x)}dx=e ^{f(x)}+c , or \int\limits_{}^{}e ^{u}(du/dx)dx = e ^{u}+c\]
anonymous
  • anonymous
\[(a) 2xe ^{x ^{2}+3}\] \[(b)(10x-2)e ^{5x ^{2}-2x}\] \[(c)(3x+2)e ^{3x ^{2}+4x+1}\] \[(d)(x ^{2}-2x)e ^{x ^{3}-3x ^{2}}\]
watchmath
  • watchmath
What have you try with these problems?

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anonymous
  • anonymous
yea i have i can do (a),(b) but c and d i got the wrong answers
watchmath
  • watchmath
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 \(\frac{1}{2}\int e^u\, du\)
anonymous
  • anonymous
oh i understand thanks but what happens to the (3x+2) inside the bracket
watchmath
  • watchmath
\((3x+2)\,dx\) become \(\frac{1}{2}du\)
anonymous
  • anonymous
ok so how about this one \[(6x ^{2}-8x+6)e ^{x ^{3}-2x ^{2}+3x-5}\]
watchmath
  • watchmath
what is the derivative of \(x^3-2x^2+3x-5\) ?
anonymous
  • anonymous
3x^2-4x+3
watchmath
  • watchmath
Good! How that derivative relates to \(6x^2-8x+6\) ?
anonymous
  • anonymous
its half of it right
anonymous
  • anonymous
the coefficients are doubled
watchmath
  • watchmath
Yes great! Then for \(f(x)=x^3-2x^2+3x-5\) the integral is \(\frac{1}{2}\int f'(x)e^{f(x)}\,dx\)
watchmath
  • watchmath
Sorry your were right :)
watchmath
  • watchmath
So \(2\int f'(x)e^{f(x)}\,dx\).
anonymous
  • anonymous
oh i see how it works out now lol:) ok thanks ill have shot at the others thankyou
watchmath
  • watchmath
Great! :D

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