5
$\begingroup$

So say for example I have $\int_{x=a}^{b}A(x)dx = \int_{x=a}^{b}B(x)dx$ for some functions $A(x)$ and $B(x)$. Then does $\int_{x=a}^{b}xA(x)dx = \int_{x=a}^{b}xB(x)dx$ still hold ? I was solving another problem and I used this without really thinking about it but now that I look at it, Im not really sure. I think it's incorrect but I just can't think of an intuitive explanation. Thanks.

$\endgroup$
0
9
$\begingroup$

One can rewrite this equality as

$$\int_{a}^{b}(A(x) - B(x)) dx= 0$$

We want to ascertain if

$$\int_{a}^{b} x(A(x) - B(x)) dx= 0$$

We can take $A(x) - B(x) = x$ if we let $a = -b$, but the second integral is of $x^2$, which is a positive function, and hence cannot be zero.

$\endgroup$
9
$\begingroup$

Not necessarily :

$$\int_0^11dx=\int_0^12xdx=1.$$ But

$\frac{1}{2}=\int_0^1xdx\not=\int_0^12x^2dx=\frac{2}{3}.$

$\endgroup$
3
$\begingroup$

Another example: $$\int_0^{2\pi}\sin x\,dx=0=\int_0^{2\pi} 0\,dx$$ but $$\int_0^{2\pi}x\sin x\,dx\ne\int_0^{2\pi} x\,0\,dx\ .$$

$\endgroup$
1
$\begingroup$

Let $$A(x)=x,\quad B(x)=-x$$ then $$\int\limits_{-1}^1 A(x)dx = 0 = \int\limits_{-1}^1 B(x)dx$$ but $$\int\limits_{-1}^1 xA(x)dx = \int\limits_{-1}^1 x^2dx = \frac 13\left[x^3\right]_{-1}^1 = \frac 23$$ $$\int\limits_{-1}^1 xB(x)dx = \int\limits_{-1}^1 -x^2dx = \frac 13\left[-x^3\right]_{-1}^1 = -\frac 23$$

Another example: $$A(x)=x,\quad B(x)=1-x$$ $$\int\limits_0^1A(x)dx=\frac 12 = \int\limits_0^1B(x)dx$$ but $$\int\limits_0^1 xA(x)dx = \int\limits_0^1 x^2 dx = \frac 13$$ $$\int\limits_0^1 xB(x)dx = \int\limits_0^1 (x-x^2)dx = \int\limits_0^1 xdx-\int\limits_0^1 x^2dx = \frac 12 - \frac 13 = \frac 16$$

edit

Yet another example. Let $f_n(x) = \delta(x-n)$ where $\delta$ is a Dirac delta function.
Then for all $n\in\{1, 2, 3, 4, 5\}$ we have $$\int\limits_0^6 f_n(x)\,dx = 1 \quad\text{but}\quad \int\limits_0^6 xf_n(x)\,dx = n$$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.