At the max/min of f(x), f ' (x) = 0 , so
?x sinx = 0
?x=0 ---> x = 0
or
sinx = 0 ---> x = 0, ?, 2?
We have to find f(x)
? ?x sinx dx ??? ---> nasty
tried integration by parts, got messy
how about this approach?
?xsinx = ? [?x(x - x^3/3! + x^4/4! - x^5/5! + ...]
using MacLaurin series
= ? [x^(3/2) - x^(7/2)/6 + x^(11/2)/24 + ...+ a constant ]
argghhh!! , have to integrate that.
ok, Wolfram to the rescue:
http://www.wolframalpha.com/input/?i=integral+%E2%88%9Ax+sinx
look at the graph of the integral, which they sketched to 2?
notice that 2? = appr 6.28
and the graph appears to have a minimum when x = 2?
Hopefully one of our other tutors will look at this and can see how to obtain the integral
f'(x)=sqrt(x)*sin(x)
The first derivative of the function f is given above. If f(0)=0, at which value of x does the function f attain it's minimum value on the closed interval [0,10]?
I know the answer is 6.28, but I need steps as to why. Please and thank you.
2 answers
forget the integral. It's not elementary.
besides, no one asked what f(x) was. The question was, where does it achieve its minimum value?
besides, no one asked what f(x) was. The question was, where does it achieve its minimum value?