2 integral problems: show int from 1 to 0 of x/(1=sinx)<1/2,

[kimmy_koo51]

Show that the integral from 1 to 0 of x/ (1=sin x) is less than 1/2

also

Let f(x)= the integral from e^2x to 2 of 1/ln t . Find f'(x)

I know this has something to do with the fundamental theorum of calculus but i am not sure how to apply it here

 

[skeeter]

Re: Please help with two integral problems I cannot figure out

Show that the integral from 1 to 0 of x/ (1=sin x) is less than 1/2

I suppose you mean x/(1+sinx) ... note that x/(1+sinx) < x/2

also

Let f(x)= the integral from e^2x to 2 of 1/ln t . Find f'(x)

I know this has something to do with the fundamental theorum of calculus but i am not sure how to apply it here

fundamental THEOREM of calculus ... is the lower limit of integration 2 or e[sup:2bvqhk6k]2x[/sup:2bvqhk6k]?

the FTC sez ... \(\displaystyle \frac{d}{dx} \int_a^u f(t) dt = f(u) \cdot \frac{du}{dx}\)

 

[Dr. Flim-Flam]

Re: Please help with two integral problems I cannot figure out

Hint: I think the answer (for the second integral) is -e^(2x)/x. I,m not going to show my work on this as I still haven't learn LaTex.

For the first integral (hint) -1 =< sin(x) <= 1.

 

[kimmy_koo51]

x/ 1+ sin x < x/2

Our possible values of x are 1 and 0 from the upper boundary and lower boundary (respectively), and therefore:

1/ 1+ sin 1 = 0.98 which is greater than 1/2, which is not consistent with our statement....

ugh, i am frustrated, is there any other possible way?

 

[Dr. Flim-Flam]

f(x) = Integral [x/(1-sin(x))],x,1,0 =integral [-x/(1-sin(x))],x,0,1 = integral x/(sin(x)-1)],x,0,1

Hence -1<= sin(x) <1, sin(x) can't equal 1 (division by zero isn't allowed), hence the denominator will always be negative,

Ergo the antiderivate will always be negative and will be < 1/2.