If there is a force $\mathbf F(x)=ax\hat i$ and an object is moving from $x_2>0$ to $x_1>0$ in the opposite direction of force. Then work could be calculated as follows

$$\int_{x_2}^{x_1} -F(x)\cdot\text dx$$

the negative sign is because the direction between $\mathbf F(x)$ and $\text dx$ are anti-parallel

$$-\int_{x_2}^{x_1} ax\cdot\text dx$$ $$-a(\frac{x_1^2-x_2^2}{2})$$

since $x_1 < x_2$, therefore $x_1^2<x_2^2$ so $$x_1^2-x_2^2 < 0$$ and $$-a(\frac{x_1^2-x_2^2}{2})>0$$

So my calculation shows that the work done in going against the force is positive, which is absolutely wrong. Please correct me!

If there is a force $\mathbf F(x)=ax\hat i$ and an object is moving from $x_2>0$ to $x_1>0$ in the opposite direction of force. Then work could be calculated as follows

$$\int_{x_2}^{x_1} -F(x)\cdot\text dx$$

the negative sign is because the direction between $\mathbf F(x)$ and $\text dx$ are anti-parallel

$$-\int_{x_2}^{x_1} ax\cdot\text dx$$ $$-a(\frac{x_1^2-x_2^2}{2})$$

since $x_1 < x_2$, therefore $x_1^2<x_2^2$ so $$x_1^2-x_2^2 < 0$$ and $$-a(\frac{x_1^2-x_2^2}{2})>0$$

So my calculation shows that the work done in going against the force is positive, which is absolutely wrong.

If there is a force $\mathbf F(x)=ax\hat i$ and an object is moving from $x_2$ to $x_1$ in the opposite direction of force. Then work could be calculated as follows

$$\int_{x_2}^{x_1} -F(x)\cdot\text dx$$

here minus is because the direction between $\mathbf F(x)$ and $\text dx$ are anti-parallel

$$-\int_{x_2}^{x_1} ax\cdot\text dx$$ $$-a(\frac{x_1^2-x_2^2}{2})$$

since $x_1 < x_2$, therefore $x_1^2<x_2^2$ so $$x_1^2-x_2^2 < 0$$ and $$-a(\frac{x_1^2-x_2^2}{2})>$$

So my calculation shows that the work done in going against the force is positive, which is absolutely wrong. Please correct me!

If there is a force $\mathbf F(x)=ax\hat i$ and an object is moving from $x_2>0$ to $x_1>0$ in the opposite direction of force. Then work could be calculated as follows

$$\int_{x_2}^{x_1} -F(x)\cdot\text dx$$

the negative sign is because the direction between $\mathbf F(x)$ and $\text dx$ are anti-parallel

$$-\int_{x_2}^{x_1} ax\cdot\text dx$$ $$-a(\frac{x_1^2-x_2^2}{2})$$

since $x_1 < x_2$, therefore $x_1^2<x_2^2$ so $$x_1^2-x_2^2 < 0$$ and $$-a(\frac{x_1^2-x_2^2}{2})>0$$

So my calculation shows that the work done in going against the force is positive, which is absolutely wrong. Please correct me!

If there is a force $\mathbf F(x)=ax\hat i$ and an object is moving from $x_2$ to $x_1$ in the opposite direction of force. Then work could be calculated as follows

$$\int_{x_2}^{x_1} -F(x)\cdot\text dx$$

here minus is because the direction between F(x) and dx are anti parallel

$$-\int_{x2}^{x1} ax.dx$$ $$-a(\frac{x1^2-x2^2}{2})$$

since x1 < x2 therefore $$x1^2<x2^2$$ so $$x1^2-x2^2 < 0$$ and $$-a(\frac{x1^2-x2^2}{2})$$ will be positive

so my calculation shows that the work done in going against the force is positive which is absolute wrong. please correct me!

If there is a force $\mathbf F(x)=ax\hat i$ and an object is moving from $x_2$ to $x_1$ in the opposite direction of force. Then work could be calculated as follows

$$\int_{x_2}^{x_1} -F(x)\cdot\text dx$$

here minus is because the direction between $\mathbf F(x)$ and $\text dx$ are anti-parallel

$$-\int_{x_2}^{x_1} ax\cdot\text dx$$ $$-a(\frac{x_1^2-x_2^2}{2})$$

since $x_1 < x_2$, therefore $x_1^2<x_2^2$ so $$x_1^2-x_2^2 < 0$$ and $$-a(\frac{x_1^2-x_2^2}{2})>$$

So my calculation shows that the work done in going against the force is positive, which is absolutely wrong. Please correct me!