Hello .:Jessica:. Go Spartans!,
I prefer to work problems like this in general terms, and then plug our given data into the resulting formula.
First, let's let:
$$\ell$$ = the length of the trough.
$$h$$ = the depth of the trough.
$$b$$ = the width of the trough at the bottom.
$$B$$ = the width of the trough at the top.
$$s$$ = the distance above the top of the trough the fluid is to be pumped.
$$\rho$$ = the weight density of the fluid.
$$g$$ = the acceleration due to gravity.
Now, let's imagine slicing the contents of the tank horizontally into rectangular sheets. The length of each sheet is constant, given by the length of the tank $\ell$. The width $w$ of each sheet will be a function of its vertical position within the trough.
So, let's orient a vertical $y$-axis paasing though the axis of symmetry of a vertical cross-section, with its origin at the bottom of the trough.
We know that the width of the trough will vary linearly as we move up the $y$-axis, and will contain the two points:
$$(0,b),\,(h,B)$$
Hence, the slope of this linear function is given by:
$$m=\frac{\Delta w}{\Delta y}=\frac{B-b}{h}$$
And so, using the slope-intercept form of a line, we find:
$$w(y)=\frac{B-b}{h}y+b$$
And so the volume of an arbitrary sheet is:
$$dV=\ell\cdot w(y)\,dy=\ell\left(\frac{B-b}{h}y+b \right)\,dy$$
Next, we want to determine the weight $\omega$ of the arbitrary sheet. Using the definition of weight density, we may state:
$$\rho=\frac{\omega}{dV}\,\therefore\,\omega=\rho\,dV$$
Next, we want to determine the distance $d$ the arbitrary sheet must be lifted. This is:$$d=s+(h-y)$$
Thus, using the fact that work is the product of the applied force and the distance through which this force is applied, we find the work done to lift the arbitrary sheet is:
$$dW=\omega d=d\rho\,dV=\rho\left(s+(h-y) \right)\left(\ell\left(\frac{B-b}{h}y+b \right)\,dy \right)=\ell\rho\left((s+h)-y) \right)\left(\frac{B-b}{h}y+b \right)\,dy$$
After a bit of algebra, we may write:
$$dW=\frac{\ell\rho}{h}\left((b-B)y^2+\left(s(B-b)+h(B-2b) \right)y+bh(h+s) \right)\,dy$$
Now, if we let $0\le h_0\le h$ be the initial depth of the fluid, we may state that the work required to pump out the contents of the trough is given by:
$$W=\frac{\ell\rho}{h}\int_0^{h_0} (b-B)y^2+\left(s(B-b)+h(B-2b) \right)y+bh(h+s)\,dy$$
Applying the FTOC, we obtain:
$$W=\frac{\ell\rho}{h}\left[\frac{(b-B)}{3}y^3+\frac{s(B-b)+h(B-2b)}{2}y^2+bh(h+s)y \right]_0^{h_0}$$
$$W=\frac{\ell\rho}{6h}\left(2(b-B)h_0^3+3\left(s(B-b)+h(B-2b) \right)h_0^2+6bh(h+s)h_0 \right)$$
In the given problem, we are told the trough is initially full, i.e. $h_0=h$, so for such a case the above formula becomes:
$$W=\frac{\ell\rho}{6}\left(2(b-B)h^2+3\left(s(B-b)+h(B-2b) \right)h+6bh(h+s) \right)$$
Expanding and collecting like terms, we obtain:
$$W=\frac{\ell\rho h}{6}\left(b(2h+3s)+B(h+3s) \right)$$
In the given problem, we are told we need only to pump the fluid to the top of the tank, i.e. $s=0$, so for such a case the above formula becomes:
$$W=\frac{\ell\rho h^2}{6}(2b+B)$$
Now, the data we are given for this problem is:
$$\ell=15\text{ m},\,\rho=\frac{1000\text{ kg}}{\text{m}^3}\cdot9.8\frac{\text{m}}{\text{s^2}}=9800\frac{\text{N}}{\text{m}^3},\,h=5\text{ m},\,b=2.5\text{ m},\,B=5\text{ m}$$
Plugging in this data, we find:
$$W=\frac{15\cdot9800\cdot5^2}{6}(2\cdot2.5+5)\text{ J}=6.125\text{ MJ}$$
