0
$\begingroup$

Okay, so basically I thought I got my answer fully correct, but seeing the correction, it seems I'm not. Either I'm wrong or the one who corrected the exam and sent the correction is. (It's a board exam and this correction was not official.)

Basically, we have a strictly increasing function $f$ and its anti-derivative $F$.

$f(k)=0$ and $f(0)=-1$.

Its inverse function is $g$.

Now, the question is to calculate:

$\int_{-1}^0g(x)dx$

Now, here is my solution:

$G$ is the symmetric to $C$ with respect to $y=x$ and the x-axis and y-axis are symmetric to each other with respect to $y=x$. Thus, the area bounded between $G$,y-axis and x-axis is the same as that bounded between $C$,y-axis and x-axis.

Thus,

$\int_{-1}^0 |g(x)|dx=\int_0^k|f(x)|dx$

$\int_{-1}^0g(x)dx=-\int_0^k f(x)dx=F(0)-F(k)$

Now, this answer is apparently wrong.

The answer adds the improper integral $\int_{-\infty}^0 (x-f(x))dx$ but writes it as a limit instead (which doesn't really change anything).

However, we NEVER took improper integrals (I know them because I read a bit of analysis) and all this seems plain wrong.

Can anyone tell me where my line of thought went wrong in my solution?

$\endgroup$
5
  • $\begingroup$ If you try this with one of the simplest possible examples, i.e., $f(x)=x-1$, which has $f(0)=-1$ and $f(1)=0$, then $\int_0^10f(x)dx=-\frac{1}{2}$. Similarly, $g(x)=x+1$ and $\int_{-1}^0g(x)dx=\frac{1}{2}$. On the other hand, $\int_{-\infty}^0(x-f(x))dx=\int_{-\infty}^0(-1)dx$, which is infinite. $\endgroup$
    – Michael Burr
    Jun 12, 2015 at 14:42
  • $\begingroup$ Hmm, okay then. Guess those who corrected it are stupid. Thanks. :) $\endgroup$
    – Hasan Saad
    Jun 12, 2015 at 14:45
  • $\begingroup$ Oh, and the improper integral was finite in this question by the way. $\endgroup$
    – Hasan Saad
    Jun 12, 2015 at 14:46
  • $\begingroup$ Well, if $f(k)$=0 then our function is clearly equal to 0 everywhere else except at the point 0. This implies $g(0)=k$ and $g(-1)=0$. $\endgroup$
    – Aleksandar
    Jun 12, 2015 at 14:46
  • $\begingroup$ What? What do you mean clearly equal to 0? $k$ is just one point... $\endgroup$
    – Hasan Saad
    Jun 12, 2015 at 14:50

1 Answer 1

Reset to default
1
$\begingroup$

I cannot see what is given to you as correction. But I think your answer is correct. Pick an $f$ satisfying the conditions and compute the integral. Check if the answer in the correction meets the result.

$\endgroup$
9
  • $\begingroup$ Can't it be SPECIFIC to this function $f$? Should I post the function? I don't know man... I feel like this country seriously has retards but I don't know... $\endgroup$
    – Hasan Saad
    Jun 12, 2015 at 14:42
  • $\begingroup$ Post the correction if possible. $\endgroup$
    – corindo
    Jun 12, 2015 at 14:47
  • $\begingroup$ Is it okay if I post it as a picture cause I received it as a picture? $\endgroup$
    – Hasan Saad
    Jun 12, 2015 at 14:50
  • $\begingroup$ if your correction is $F(0) - F(k) + \int_{-\infty}^0(x-f(x))dx$ then it is definitely wrong because no condition on $f$ guarantees that the last integral exist. $\endgroup$
    – corindo
    Jun 12, 2015 at 14:53
  • $\begingroup$ It does exist in this case and it is equal to $-1$. Maybe it doesn't exist in other cases, but it does exist here, and I really cannot find the reason why it'd be put in here... $\endgroup$
    – Hasan Saad
    Jun 12, 2015 at 14:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .