How Does the Integral of 1/(4y-1) with Respect to x Evaluate from 0 to 1?

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The integral of 1/(4y-1) with respect to x from 0 to 1 evaluates to x/(4y-1) after recognizing that y is treated as a constant during integration. The discussion highlights confusion regarding the variable of integration, clarifying that the integral is indeed with respect to x, not y. This distinction is crucial for correctly applying integration techniques in calculus.

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integral of 1/(4y-1) dx from 0 to 1

im stumped
 
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nick727kcin said:
integral of 1/(4y-1) dx from 0 to 1

im stumped

Whats the integral of 1/y ?

Ohh I missed what LocationX pointed out, are you sure it's a function of y that you are integrating with respect to x? Or is the y supposed to be x, or the dx dy?
 
Last edited:
if it is with respects to x, then the integral is just [tex]\frac{x}{4y-1}[/tex]
 

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