Your procedure is correct, but you have made a mistake with the sign convention. Apparently you are using the same convention as I do, where a vertical load/force is negative, and an upward force is positive.
Consequently, as you've written correctly $$ w(x)=\frac{w_0}{L}x-w_0 $$
Your mistake happens as you formulate the force equilibrium equation. With this definition of $w(x)$ you already comply to the sign convention. If you now formulate the shear force equation and write $$ V_1=15-\int w(x)dx $$ you basically reverse the sign convention again. To formulate the force equilibrium equation you have to sum all forces, not subtract them, thus $$ V_1=15+\int w(x)dx $$ which leads to $$ V_1=15+\frac{5}{3}x^2-10x $$ which is the correct result.
Sign convention [edited]
Take a look at your $w(x)$. It's a force pointing downwards, so it should be negative. You have written it as $$ w(x)=\frac{w_0}{L}x-w_0 \qquad \mbox{for}\qquad x={0...3}$$ Thus $w(0)=-w_0$, which means your load $w(x)$ is already defined in the coordinate system you specified. 7
If you now sum or integrate and add a minus sign in front of $w(x)$ you basically turn the downward load into an upwards facing load.
Consider, e.g. $w(x)=const.=-w_0$, i.e. an evenly distributed downwards load, thus a negative value. It's easy to figure out, that the reaction at the bearing is $A_y=\frac{1}{2} w_0L $. Now to find the shear force distribution, if we use your method, we'd write: $$ V_1=A_y-\int_0^L (-w_0)dx=\frac{1}{2} w_0L - (-w_0)L=\frac{3}{2}w_0L $$ This is the wrong result, because for the shear force distribution calculation we suddenly switch sign convention (subtracting during integration instead of adding).
In the first part you said $w(x)$ was $w(x)=-w_0+\frac{x}{L}w_0$, which goes linearly from $-w_0$ to $0$, which in your example would be $w(x)=\frac{10}{3}(x-3)$. During your second calculation you managed to get the right result because you changed the sign convention of your $w(x)$, as you inserted $w(x)=\frac{10}{3}(3-x)$ (which is an upwards load), but then you turned it downwards again by adding a negative sign in front of the whole term.
I really recommend sticking to one sign convention. Either you say you sum all forces, and defined downward loads as negative, or you subtract downward loads from upward reactions. 7