Calc 3: Calculate Work Done on Particle I've been working on this problem for a while and I'm pretty stuck. I tried it multiple different ways, by the last time I attempted it I realized that I hadn't converted kilometers to meters the entire time.
Anyway, the last approach I tried after doing the conversion was to integrate over the function from 7,000,000 to 8,300,000. I got a positive answer back, which I didn't think made sense given that the force would be opposing the movement, so I flipped signs. After integrating, I then multiplied the answer to the displacement (8,300,000 - 7,000,000) to get my final answer. 
My logic was that integrating over the function would give me the total force, multiplying by the displacement would give me that total force over the distance which was the work. I don't understand what's wrong with this solution. This is a homework problem but at this point I've gotten it wrong too many times to get any points for it, I'm just itching to know how it's done.
Picture of the problem:

Smart Method
The work done is equal to the change in potential energy. The potential of a gravitational field:
$$
v(\vec{r})=-\frac{GM}{\|\vec{r} \|}\ \phantom{aaaa}\dots(1)
$$
So the work done is
$$
m\Delta v=-\frac{GMm}{\|\vec{r}_1 \|}+\frac{GMm}{\|\vec{r}_2 \|}
$$
Where $\vec{r}_1$ is the initial radial vector and $\vec{r}_2$ the final.
This works as the potential is defined as the work done on a particle of unit mass in bringing it in from infinity to its final radial distance (which is why the sign in the potential is negative as the particle does work in moving from infinity). You can derive $(1)$ above by integrating force equation given in the question if you so desire.
Your answer in the attachment is wrong as you have the radial distances squared. 
The gist of this can be found on the Wikipedia page on Gravitational Potential
Brute Force and Ignorance method:
$$
W=\int_{r=r_1}^{r_2}F\;dr=\int_{r=r_1}^{r_2}-\;\frac{GMm}{r^2}\;dr=\frac{GMm}{r_2}-\frac{GMm}{r_1}
$$
