What is Fourier transform in terms of area under the curve?
Since

$$X(\omega)=\int_{-\infty}^{\infty}x(t)e^{-j\omega t}dt$$

you get for the real part of $X(\omega)$ (assuming that $x(t)$ is real)

$$X_R(\omega)=\int_{-\infty}^{\infty}x(t)\cos(\omega t)dt\tag{1}$$

and for the imaginary part

$$X_I(\omega)=-\int_{-\infty}^{\infty}x(t)\sin(\omega t)dt\tag{2}$$

So if you like, from (1) and (2) you can interpret the real part of the Fourier transform as the area under the curve $x(t)\cos(\omega t)$, and the imaginary part would then be the area under the curve $-x(t)\sin(\omega t)$.

Note that interpreting an integral as the area under a curve does not always help intuition. In the case of the Fourier transform it is more natural to interpret the integrals (1) and (2) as inner products.