By Green's theorem, the line integral
[tex]\displaystyle \int_C f(x,y)\,\mathrm dx + g(x,y)\,\mathrm dy[/tex]
is equivalent to the double integral
[tex]\displaystyle \iint_D \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \,\mathrm dx\,\mathrm dy[/tex]
where D is the region bounded by the curve C, provided that this integrand has no singularities anywhere within D or on its boundary.
It's a bit difficult to make out what your integral should say, but I'd hazard a guess of
[tex]\displaystyle \int_C \left(3y+5e^{-x}\right)\,\mathrm dx + \left(10x+3\cos\left(y^2\right)\right)\,\mathrm dy[/tex]
Then the region D is
D = {(x, y) : 0 ≤ x ≤ 1 and x ² ≤ y ≤ √x}
so the line integral is equal to
[tex]\displaystyle \int_0^1\int_{x^2}^{\sqrt x} \frac{\partial\left(10x+3\cos\left(y^2\right)\right)}{\partial x} - \frac{\partial\left(3y+5e^{-x}\right)}{\partial y}\,\mathrm dy\,\mathrm dx \\\\ = \int_0^1 \int_{x^2}^{\sqrt x} (10-3)\,\mathrm dy\,\mathrm dx \\\\ = 7\int_0^1 \int_{x^2}^{\sqrt x} \mathrm dy\,\mathrm dx[/tex]
which in this case is 7 times the area of D.
The remaining integral is trivial:
[tex]\displaystyle 7\int_0^1\int_{x^2}^{\sqrt x}\mathrm dy\,\mathrm dx = 7\int_0^1y\bigg|_{y=x^2}^{y=\sqrt x}\,\mathrm dx \\\\ = 7 \int_0^1\left(\sqrt x-x^2\right)\,\mathrm dx \\\\ = 7 \left(\frac23x^{3/2}-\frac13x^3\right)\bigg|_{x=0}^{x=1} = 7\left(\frac23-\frac13\right) = \boxed{\frac73}[/tex]
