Confusion in Line integral

The value of the integral of the function $g(x, y) = 4x^3 + 10y^4$ along the straight line segment from the point $(0, 0)$ to the point $(1, 2)$ in the x-y plane is.

$(A)33$ $(B)35$ $(C)40$ $(D)56$

I have Confusion in the line integral. the thing is which method is correct.

Method 1

From this, we get value in the option

The equation of the straight line from $(0, 0)$ to $(1, 2)$ is $y = 2x$. Now, $g(x, y) = 4x^3 + 10y^4$.

Or, $g(x, 2x) = 4x^3+ 160x^4$

Now $\displaystyle \int_0^1 g(x, 2x) = \int_0^1 (4x^3 + 160x^4) ~ dx = 33$

Method 2

This is the case of scalar line integral. so $$\displaystyle \int_c f(x, y) = \int_a^b f(x(t),y(t))|r'(t)|~ dt $$ $$r(t)=\lt 0,0\gt + t<1,2> =<t,2t> ; 0\le t \le 1$$

Thus, $ r'(t)=<1,2> and |r'(t)|=\sqrt {1^2+2^2}=\sqrt {5}; ds=|r'(t)|dt=\sqrt {5}dt $

$$\displaystyle \int g(x, y)~ds = \int_0^1 (4t^3+10(2t)^4)\sqrt{5}~ dt $$ $$=\sqrt{5}\times33$$ $$=73.8$$

so which one is wrong.and where i am thinking wrong?plz suggest.

Your first method is incorrect. The question seeks you to find integral of the function $g(x, y)$ along the straight line from $(0, 0)$ to $(1, 2)$. In other words, you are to find

$ \displaystyle \int_C g(x, y) ~ ds$.

Note that $ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} ~ dx = \sqrt5 ~dx$