integration help

[hot hemlock]

For the first qs, I don’t understand why we integrate the functions with respect to x using x values as bounds.

I thought when it says that it is bounded by the y-axis, we use the y values as bounds and evaluate the integral with respect to y. idk im confused

[warm canopy]

i think you can use integration wrt y its just that its convention to take wrt x

or you can just convert the functions into the form x=f(y)

and integrate wrt y

because logarithmic and exponential functions are one to one

[hot hemlock]

so it doesn’t matter in this case?

[hot hemlock]

does this apply to every area using integration

[warm canopy]

nope

but with other functions that arent one to one you cant use that

[warm canopy]

not every, sometimes you cant find the area wrt y because the curve can not be expressed as a function in terms of y

for example most polynomials in x w degree > 1 cant be expressed in terms of y

[hot hemlock]

ahh ok. So it doesn’t matter whether u integrate wrt y or x for one to one functions, but it does for certain polynomials

[daring dune]

It would not make sense to integrate a function with respect to x with something else than bounds that live on the x axis
Because the integration variable is x, so you have to express the bounds of your integral with respect to this integration variable

As Renoir said the name of the variable does not matter when doing the computation, but when you have x-axis and y-axis it is confusing to use y as the variable living on the x-axis

[warped whale]

inetgral means adding up small pieces, u should interpret it as this instead of only area under the curve

think of integral as summation (its means that only)

but continous ver

and the expression $$\int f(x) dx \text{means adding rectangles with height f(x) and width dx over the bounds}$$

[civic lynx]

Ɱιყυ

[warped whale]

so u r adding rectangles in the x direction

u can also add both dx and dy

which is the doouble integral

with height dy and width dx

with bound e(-x - 1/4 to e^-2x in y direction

and 0 to ln2 in x direction

and both will give same answer

[warm canopy]

I meant that both are possible it's just that depending on the bounds you'll get different answers