anonymous
  • anonymous
cal 1 question
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
okay so you can separate the integral like \[\int\limits_{-5}^{1}f(x) dx= \int\limits_{-5}^{0}82xdx+ \int\limits_{0}^{1}6x^2dx\]
anonymous
  • anonymous
technically you would have to evaluate the lower limit of the integral of the second term ( the 0 in the 6x^2 integral ) at some letter (a,b,c ect what ever letter you like). then you take the limit of the letter as it approaches the zero that is there as you see it above.

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anonymous
  • anonymous
So in a more punctual approach the integral should be \[ \int\limits_{1}^{-5}f(x) dx = \int\limits_{-5}^{0}82x dx + \lim_{a \rightarrow 0} \int\limits_{a}^{1}6x^2dx\]
anonymous
  • anonymous
It's kind of technicality you will definitely see in later calculus called improper integrals. The REASON for this is because of the inequality \[x > 0\] which means the x on that interval never actually reaches zero, so you say as x --> 0 (as x approaches zero, 0.00000001 and smaller). Now just take the integral of either one, the second one is the most correct if you understand the limit deal. but both will give you the same answer.
anonymous
  • anonymous
then how do u do it with that limit?
anonymous
  • anonymous
you would evaluate the integral first then take the limit. so \[\lim_{a \rightarrow 0} \int\limits_{a}^{1} 6x^2dx = \lim_{a \rightarrow 0} 6x^3/3- a \rightarrow1\]
anonymous
  • anonymous
then \[6(1)^3/3 - \lim_{a \rightarrow 0}6(a)^3/3 = 2(1)^3 -\lim_{a \rightarrow 0}2(a)^3 = 2 - 0 = 2\]
anonymous
  • anonymous
But that's not the correct answer to it!
anonymous
  • anonymous
thats only half of it though. you have to take the integral of the first half also, i was only showing how to do the limit part.
anonymous
  • anonymous
do you see? so evaluate \[\int\limits_{-5}^{0}82xdx\] and add two
anonymous
  • anonymous
answer should be 1027?
anonymous
  • anonymous
Nope :(
anonymous
  • anonymous
what is the answer? is it -1023?
anonymous
  • anonymous
Not even that..
anonymous
  • anonymous
answer is?
anonymous
  • anonymous
I can just type in the possible answers to see if thats right or wrong.. It doesn't tell the exact ans though
anonymous
  • anonymous
okay
anonymous
  • anonymous
did either work?
anonymous
  • anonymous
no

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