anonymous
  • anonymous
Integration
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[Given~\int\limits_{k}^{2}(3-4x)dx=3,find~the~value~of~k\]
anonymous
  • anonymous
This is my working...\[\left[ 3x-2x^2 \right]_{k}^{2}=3\]\[\left[ 3(2)-2(2)^2)-(3(k)-2(k)^2 \right]=3\]\[2k^2-3k-5=0\]\[k=\frac{ 5 }{ 2 },k=-1\]
anonymous
  • anonymous
yea correct

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anonymous
  • anonymous
But the book's answer say is k=-1
anonymous
  • anonymous
can anyone explain why is it k=-1 not k=-1 and k=5/2 ?
anonymous
  • anonymous
is it given that the solutions must be in integer form?
amistre64
  • amistre64
|dw:1447509340369:dw|
amistre64
  • amistre64
5/2 is the result of a > b ... by convention, the upper limit higher than the lower limit. and if they are switched about we get \[\int_{b}^{a}=-\int_{a}^{b}\]
amistre64
  • amistre64
\[\int_{2}^{5/2}f(x)~dx=-3\] \[\int_{5/2}^{2}f(x)~dx=3\]
dayakar
  • dayakar
(k+1)(2k-5)=0 here we can write k+1=0 or 2k-5=0 k=-1 or k= 5/2 in your problem they took k=-1 u can either value
amistre64
  • amistre64
it is most likely that the book is assuming that 2 is the upper limit of the domain, so anything greater than 2 is extraneous
anonymous
  • anonymous
Thnx @amistre64 @at0m @dayakar
anonymous
  • anonymous
np B)

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