Area Under Cartesian Curve:

The area of the region enclosed by a curve y = ƒ(x), the axis of x, and two coordinates x = a and x = b is

∫^b_a  ydx.

Proof: Let P (x, y), Q (x + Δx, y + Δy) be two neighboring points on the provided curve such that the functions y = ƒ(x) is decreasing or increasing in the interval [x, x + Δx]. Point the coordinates PL and QM. Complete the rectangles LQ and MP. Take that

S = Area of the section CLPA (S is a function of x)

And S + ΔS = Area of the section CMQA.

∴ ΔS = Area of the section LMQP.

We have LM = OM - OL = x + Δx - x = Δx.

Undoubtedly, Area of the rectangle MP < ΔS < Area of the rectangle LQ

=> y Δx < ΔS < (y + Δy) Δx

=> y < ΔS/Δx < y + Δy.                                  (1)

Let Q-> P so that Δx->0. Then from (1) we get

dS/dx = y.

∴ ∫^b_a  ydx = ∫^b_a  dS/dx dx = |S|^b_a

= [S]x = b - [S]x = a

= Area of the section ABDC - 0.

Hence the necessary area =  ∫^b_a  ydx.

Remark: Similarly it may be proved that:

The area of the region enclosed by a curve x = Ø(y), the axis of y, and the two lines y = c and y = d is

∫^d_c  xdy.