Integration Q Homework: Integrate 4 with No Respective Variable?

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Homework Statement



If ∫4x dx is read as integrate 4 with respect to x, then what is there's no x so that ∫4 d?. Integrate 4 with respect to nothing?
 
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studentxlol,

∫4x dx is read, "the integral of 4x with respect to x."

Underhill
 
studentxlol said:

Homework Statement



If ∫4x dx is read as integrate 4 with respect to x,
No, it isn't. It would be read as "integrate 4x with respect to x"

then what is there's no x so that ∫4 d?. Integrate 4 with respect to nothing?
\int 4 d is an incorrect notation. You do not need an x in the function to be integrated (it can be a constant function) but you cannot have "d" in an integral without a variable: \int 4dx= 4x+ C, \int 4 dt= 4t+ C, etc.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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