# If I have an integral equality and then I multiply a variable inside the integral, does the equality still hold?

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.

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.

Not necessarily :

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

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

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

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 $$\int\limits_0^6 f_n(x)\,dx = 1 \quad\text{but}\quad \int\limits_0^6 xf_n(x)\,dx = n$$