Math2400
  • Math2400
can someone explain this derivative ? i have the answer but the steps are unclear to me
Mathematics
chestercat
  • chestercat
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Math2400
  • Math2400
\[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\]
amistre64
  • amistre64
what is the fundamental thrm of calculus for integration?
Math2400
  • Math2400
\[ \int\limits_a^b f(x)dx=F(b)-F(a)\]

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amistre64
  • amistre64
good, but lets use different notation, f= f', and F = f this helps to see the relationship between F and f in a more normal fashion. \[\int_{a}^{b}f'(x)~dx=f(b)-f(a)\] same setup, just more familiar notations what happens when we take the derivative of both sides?
Math2400
  • Math2400
okay got that, and um you get the derivative..? lol
amistre64
  • amistre64
whats it look like tho, we are working a generalization
Math2400
  • Math2400
don't u have to undo the antiderivative first on the Right side?
Math2400
  • Math2400
*left
amistre64
  • amistre64
we already have, its the right side
amistre64
  • amistre64
the right side is the 'undoing' of the derivative of the left side. f' comes from f
Math2400
  • Math2400
oh okay so if i take the derivative of f(b)-f(a)?
amistre64
  • amistre64
yep
amistre64
  • amistre64
almost, a and b need not be constants, they can be functions, so we apply the chain rule \[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=\frac d{dx}f(b)-\frac d{dx}f(a)\] \[\frac d{dx}\left[\int_{a}^{b}f'(t)~dt\right]=b'~f'(b)-a'~f'(a)\]
amistre64
  • amistre64
notice here, that we already know what f' is ... we know a and b, therefore we know a' and b'
Math2400
  • Math2400
ok so u u just need to get the derivatives of a and b and multiy by f'
amistre64
  • amistre64
\[\frac{ d }{ dx } [\int\limits_{0}^{x^5} \cos(3t+1) dt ]\] f'(a) = cos(3(0)+1) f'(b) = cos(3x^5+1) a(x) = 0, a' = 0 b(x) = x^5, b' = 5x^4
amistre64
  • amistre64
you have all the parts required to determine the results, without havig to intgrate first, and then take the derivative ... its a shortcut if you will
Math2400
  • Math2400
so when i re-write it into the equ. i use f'(a) and f'(b) instead of putting the original....so there should be no "t" variable
amistre64
  • amistre64
correct, the a and b are in the domain of t in this case.
Math2400
  • Math2400
i got it actually!
Math2400
  • Math2400
thank u so much :)
amistre64
  • amistre64
youre welcome

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