1
$\begingroup$

I've the following integral, which should result in 1, as shown by the scetch, but in my calculation I get the result 0. What's my mistake?

Sorry the comments are in German and please note that a German 1 often looks like an English 7. Anything in the picture which looks like a 7 to you is in fact a 1.

enter image description here

$\endgroup$
1
  • $\begingroup$ I'm actually dealing with a more complex formula, but the above formulas are what it boils down to, and which seem to have the same mistake as my more complex formulas have. $\endgroup$
    – Daniel S.
    May 27, 2015 at 19:35

2 Answers 2

Reset to default
1
$\begingroup$

The issue is you didn't change your differential. $dz = -dx$ fixes it. Your function is even so you could have simply worked with the integral from $0$ to $1$ instead.

$\endgroup$
7
  • $\begingroup$ Alternatively, you can compute the area using triangles if you are so inclined. $\endgroup$
    – Cameron Williams
    May 27, 2015 at 19:40
  • $\begingroup$ Or notice that you just have two halves of a unit square. Or... $\endgroup$
    – user137731
    May 27, 2015 at 19:41
  • $\begingroup$ How does dz look like if z(x) is more complex, let's say z(x) = 5x^2? $\endgroup$
    – Daniel S.
    May 27, 2015 at 19:41
  • $\begingroup$ @Bye_World That's another good observation! $\endgroup$
    – Cameron Williams
    May 27, 2015 at 19:41
  • $\begingroup$ @DanielS. In that case, $dz = 10 x\,dx$. You take the derivative of $z$ and get that $dz = z'(x)\,dx$. $\endgroup$
    – Cameron Williams
    May 27, 2015 at 19:42
0
$\begingroup$

The issue is that you changed the bounds wrong. You correctly wrote $$ \int_{z(-1)}^{z(0)} $$ and then incorrectly changed it to $$ \int_{0}^{1} $$ when it should be $$ \int_{1}^{0} $$

One final note though. You should not use substitution to deal with a multiplicative constant. Just use linearity: $$ \int_{a}^{b} c f(x)dx = c \int_{a}^b f(x)dx $$

The $-1$ in your problem you could just "pull out" of the integral.

$\endgroup$
4
  • $\begingroup$ I cannot see where I mistakenly swap the bounds. I'm mostly modifying the first of the pair of integrals only. The second one stays with the bounds 0,1 for some time while the first changes from z(-1),z(0) to correctly 1,0 and only later to 0,1 while I also attach a minus. $\endgroup$
    – Daniel S.
    May 27, 2015 at 19:36
  • 1
    $\begingroup$ You missed the negative sign in his work. He didn't make the mistake when switching the bounds. It was just that he omitted $dz = -dx$. $\endgroup$
    – Cameron Williams
    May 27, 2015 at 19:39
  • $\begingroup$ Oh I assumed the $-$ sign in front of his integral was from $dz = -dx$ and not from switching the bounds. I'll leave my answer since it's not clear to me which of the two mistakes he made. $\endgroup$
    – nullUser
    May 27, 2015 at 19:40
  • $\begingroup$ The OP made the substition $z = -x$, but didn't correctly substitute $-dz = dx$ $\endgroup$
    – MathNewbie
    May 27, 2015 at 19:41

You must log in to answer this question.

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