anonymous
  • anonymous
f(x)=definite integral 1/(4+e^t)dt, [0, ln x], with x>0. find (a) f(1) (b) f'(1) (c) inverse of f(0)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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IrishBoy123
  • IrishBoy123
\[f(x)=\int\limits_{t=0}^{\ln x} dt \qquad \dfrac{1}{4+e^t} \]
jim_thompson5910
  • jim_thompson5910
Hint for part (a) \[\Large f(x)=\int\limits_{t=0}^{t=\ln x} \dfrac{1}{4+e^t}dt\] \[\Large f({\color{red}{x}})=\int\limits_{t=0}^{t=\ln ({\color{red}{x}})} \dfrac{1}{4+e^t}dt\] \[\Large f({\color{red}{1}})=\int\limits_{t=0}^{t=\ln ({\color{red}{1}})} \dfrac{1}{4+e^t}dt\]
anonymous
  • anonymous
so that mean that f(1) = 0

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jim_thompson5910
  • jim_thompson5910
why is f(1) equal to 0?
jim_thompson5910
  • jim_thompson5910
you are correct, but I want to make sure you are using the right integral property
anonymous
  • anonymous
it is because The Second Fundamental Theorem of Calculus If ƒ is continuous on an open interval I containing a, then for every x in the interval, ƒ(t) dt = ƒ(x)
jim_thompson5910
  • jim_thompson5910
no, it's because of this property \[\Large \int_{a}^{a}f(x)dx = 0\]
jim_thompson5910
  • jim_thompson5910
the area under the curve from x = a to x = a is going to be 0 because the width of this region is 0 units wide
jim_thompson5910
  • jim_thompson5910
think of an infinitely thin strip or line with 0 area
anonymous
  • anonymous
but it is going from x=o to x= ln(1)
jim_thompson5910
  • jim_thompson5910
and ln(1) = 0
jim_thompson5910
  • jim_thompson5910
\[\Large \log_b(1) = 0\] for any valid base b
anonymous
  • anonymous
yes , you're right
jim_thompson5910
  • jim_thompson5910
so \[\Large f({\color{black}{1}})=\int\limits_{t=0}^{t=\ln ({\color{black}{1}})} \dfrac{1}{4+e^t}dt\] turns into \[\Large f({\color{black}{1}})=\int\limits_{t=0}^{t=0} \dfrac{1}{4+e^t}dt\]
jim_thompson5910
  • jim_thompson5910
for part (b), you'll need to use the fundamental theorem of calculus http://aleph0.clarku.edu/~ma120/FTC.jpg the second part of it. Don't forget to use the chain rule
anonymous
  • anonymous
for part b, is it also 0
jim_thompson5910
  • jim_thompson5910
tell me what you got for \(\Large f \ '(x)\)
jim_thompson5910
  • jim_thompson5910
this page may help when it comes to the chain rule: http://www.sosmath.com/calculus/integ/integ03/integ03.html
jim_thompson5910
  • jim_thompson5910
Look at the part that has \[\LARGE F(x) = \int_{a}^{u(x)}f(t)dt\]
jim_thompson5910
  • jim_thompson5910
notice just under that is \[\LARGE F \ '(x) = f{\Huge(}u(x){\Huge)}u \ '(x)\]
jim_thompson5910
  • jim_thompson5910
`for part b, is it also 0` no, the answer to part (b) is NOT zero
anonymous
  • anonymous
so part b is 1/5
jim_thompson5910
  • jim_thompson5910
how did you get 1/5
jim_thompson5910
  • jim_thompson5910
what f ' (x) function did you get?

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