# Why can't I do this?

1. Mar 16, 2004

### DavidWi

Ok, I was thinking today during my calculus class about taking the integral of a function in a different way. Let's assume for a second that we want to find the area between the function and the y axis, on the interval x = [0, 2] of the function y = 2x.

What I was thinking I could do, is take the integral of y.

(y^2)/2 then substitute 2x in for y, since y = 2x.

(2x)^2 / 2

That would give us the integral between the function and the y axis and we would be able to put in the interval for x.

But it didn't work... I got the wrong answer? Why doesn't this work? I think it should work.

Thanks.

2. Mar 16, 2004

### matt grime

Because that isn't how integration works?

Imagine doing the same for differentiation.

we want to find d/dx of x^2, well, d/dy of y is 1, so putting y=x^2, we get d/dx(x^2)=1

It just isn't right.

More formally remember integration is wrt something

so integral of ydy is not the same as integral of x^2dx when you put x^2=y because the dy and dx are there, and if y=x^2, then dy is not dx - it is 2xdx

3. Mar 17, 2004

### HallsofIvy

y=2x and x ranges between 0 and 2.

Of course, that's simply a right triangle with base (x-axis) of length 2 and height (y-axis) of length 4: its area is (1/2)(2)(4)= 4.

You could do this as $$\int_0^2(ydx)= \int_0^2(2x)dx= x^2\|_0^2= 4$$

You could do this as $$\int_0^4(xdy)= \int_0^4\frac{y}{2}dy= \frac{y^2}{4}\|_0^4= \frac{16}{4}= 4$$

Your formula is wrong because you never took into account the "dx" or "dy".

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