0
$\begingroup$

I'm having trouble with the following two questions, one of which i don't know how to solve totally (both are among a sub questions of a bigger question).

a) prove that there is $c\in [0,1]$ so that $\int _0^1\:\sin (x^3)dx=\int _0^c\sin\left(x^2\right)dx$ (i've tried to compute the integrals and then compare them, and to assert that using the mean value theorem $$ \int_{a}^{b} f (t)dt = f (c) \cdot (b - a)$$ there must be $c \in [0,1]$ that solves that equation. But I can not write a proof for that.

b) $f,g,h$ are bounded functions(i don't know the mathmatical terms for it) in [a,b] so for every $x \in [a,b]$, with $f(x)\le g(x)\le h(x)$. assume that $f$ and $h$ are integrable in $[a,b]$ and $\int _a^b\:f\left(x\right)dx\:=\:\int _a^b\:h\left(x\right)dx\:$. prove that $g$ is also integrable in $[a,b]$ (i don't know how to show that. But what i understand is that $f(x)$ and $h(x)$ are continuous as it's a neccesity in order for them to be integrable, so using the mean value theorem we can show $g(x)$ is integrable as well, but i don't know how to).

I hope that you can help me with those as i'm totally stuck. Thank you very much for your help and i tried to elaborate as much as i can.

$\endgroup$
  • 3
    $\begingroup$ $x^3 < x^2$ on $(0,1)$. Hence, $\sin(x^3) < \sin(x^2)$ there. This gives $A := \int_0^1\sin(x^3)\,dx < \int_0^1\sin(x^2)\,dx$. Now, $f(x) := \int_0^x\sin(t^2)\,dt$ is continuous and $f(0) < A < f(1)$. By the intermediate value theorem, there must be a $c\in (0,1)$ such that $f(c) = A$. $\endgroup$ – amsmath Oct 31 '17 at 15:25
  • 1
    $\begingroup$ Your "question" (b) is completely non-understandable. What are $f,g,h$? What is a "blocked function"? What is your question? $\endgroup$ – amsmath Oct 31 '17 at 15:30
  • $\begingroup$ thank you for replying and helping me with a). regarding b: by blocked i meant *bounded*(my english is not so good). in b i was asked to show that g(x) is also integrable in [a,b]. $\endgroup$ – BeginningMath Oct 31 '17 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.