anonymous
  • anonymous
\[\LARGE f'(x)=x^2+3x \;\; ,\;\; f(1)=5\;\;\;\;\Longrightarrow f(2)=?\] I need some hints
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
f(x) can be obtained by integration........then use f(1) to find value of c, then find f(2)
anonymous
  • anonymous
ya suroj ur correctt
anonymous
  • anonymous
got it?

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anonymous
  • anonymous
\[\LARGE f(x)=\frac13 x^3 +\frac32 x^2 \;\;\;\;??\]
anonymous
  • anonymous
you left +C at the end
anonymous
  • anonymous
you need to add +C at end because it is integration
anonymous
  • anonymous
indefinite integration
anonymous
  • anonymous
ok but... I need to firnd f(2) then how??
anonymous
  • anonymous
on the second condition when x=1, y=5, you see , put those values and get c
anonymous
  • anonymous
\[\LARGE \int\limits x^2+3x \;dx = \frac13 x^3+\frac32x^2+C\]
anonymous
  • anonymous
put the c back on f(x) and get complete f(x) equation and then find f(2)
anonymous
  • anonymous
yes
anonymous
  • anonymous
correct
anonymous
  • anonymous
how can I know what the value of C is ??
anonymous
  • anonymous
I see it you told me before... let me try, I'll see what I can do :D
anonymous
  • anonymous
see second condition f(1) is 5 than mean when x=1 , y=5
anonymous
  • anonymous
see f(x) is what you get by integration
anonymous
  • anonymous
because you integrate f'(x)
anonymous
  • anonymous
did you catch it?
anonymous
  • anonymous
is C=19/6 ??
anonymous
  • anonymous
I'm still dizzy ! O_O
anonymous
  • anonymous
yes you got it
anonymous
  • anonymous
now put that C back in f(x)
anonymous
  • anonymous
got it?
anonymous
  • anonymous
ahh yeah... now I'll do it thank you very much ;)
anonymous
  • anonymous
no problem lol
anonymous
  • anonymous
\[f \left( x \right) = x ^{2} + 3x\] \[f \left( 1 \right) = 1^{2} + 3 \times 1 = 5\] \[f \left( 2 \right) = 2^{2} + 3 \times 2 = 10\] simple.
anonymous
  • anonymous
@kreshnik another simple way to do something.
anonymous
  • anonymous
@JoshDavoll WRONG... the answer IS \[\frac{71}{6}\] !! Thanks for the help !
anonymous
  • anonymous
hahaha... obviously , you mus be 10-12 years ... LOL

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