Integration-finding an upper limit of integration. I'm given the value of an integral: $$\int_1^a (x-\frac{w^3}{x^2})\,dx=1.5$$ and told that the total area between the graph of $$y=x-\frac{w^3}{x^2}$$ and the x-axis is 6.5 units squared. w and a are both integers greater than 1. How can I find the values of w and a?
 A: $$\int_1^a \left(x-\frac{w^3}{x^2}\right) \, dx=\left[\frac{w^3}{x}+\frac{x^2}{2}\right]_1^a=\frac{a^2}{2}+\frac{w^3}{a}-w^3-\frac{1}{2}=\frac{3}{2}$$
The second condition is unclear. The graph of the family of functions $$y=x-\frac{w^3}{x^2}$$
is like the one in the image below. There is no finite area.
When OP will tell what is the area we are talking about I'll try to help further.

$$...$$

A: Let $f(x)=x-\frac{w^3}{x^2}$.
We know that the function is increasing on $(0,\infty)$ and that it has $x$-intercept $(w,0)$. So we have
$$ \int_1^w f(x)\,dx+\int_w^a f(x)\,dx=1.5\tag{1} $$
$$ -\int_1^w f(x)\,dx+\int_w^a f(x)\,dx=6.5\tag{2} $$
Would you like me to work it further or do you see how to proceed?
Here is the next step:
Subtract equation (2) from equation (1) and divide by 2 to get
$$ \int_1^w f(x)\,dx =-2.5\tag{3} $$
but you can also find the value of the integral directly
$$ \int_1^w f(x)\,dx =-w^3+\frac{3}{2}w^2-\frac{1}{2}\tag{4} $$
giving
$$ 2w^3-3w^2-4=0 $$
which has a real solution $w=2$.
Now if you add equations (1) and (2) and divide by $2$ you will get
$$ \int_2^a f(x)\,dx=4 $$
If you evaluate this integral directly you will get an equation which you can solve for $a$, which is also an integer.

