# Evaluate the line integral of a parabola

How can I evaluate :

$$\int_{C} y \;dx + x^2 \; dy$$

where $C$ is the parabola define by

$$y=4x-x^2 \quad \text{from } \; (4,0) \; \text{ to } \; (1,3).$$

Do I need to parameterize the parabola?

• Typically you should indicate the parameterization used...in this case there is a natural one which is to use x as the parameter Mar 16, 2016 at 19:24
• Note the orientation of the curve as well. Mar 16, 2016 at 19:35
• Ok... So, how can I do that? Mar 16, 2016 at 19:39

$$\because y = 4x-x^2 , \therefore \frac{dy}{dx} = 4-2x , \therefore dy = (4-2x)dx$$ Hence, $$\int_{C} y \;dx + x^2 \; dy = \int_{4}^{1} 4x-x^2 dx + \int_{4}^{1} x^2 (4-2x)dx$$ $$\implies \int_4^1 4x-x^2+4x^2-2x^3$$ $$\implies \int_4^1 4x+3x^2-2x^3$$ $$\implies (2x^2+x^3-\frac{x^4}{2})_4^1$$ $$\implies (2+1-\frac{1}{2}) - (32+64-128)$$ $$\implies 35-\frac{1}{2}$$ $$\implies \frac{69}{2}$$
• Isn't $\int_{4}^{1} x^2 \cdot (4-2x) \; dx$? Mar 16, 2016 at 21:48
The points on the parabola are $P=(x,y)=(x,4x-x^2)$ for $4\ge x\ge 1$ and $$dy=\frac{dy}{dx}dx=(4-2x)dx$$