Bidding Item Problem 
You are bidding on an item that has an unknown value uniformly
  distributed between 0 and 1. You do not know the true value of the
  item, but you know that if you end up winning the bid for the item,
  the item will increase its value to 2x its original value. Your bid
  can only go through if its at least as large as the original value of
  the item. How do you bid to maximize expected payoff.

Here's what I have: 
Let V be the true value of the item
Let B be the bid you make
Let f(V) represent the profit you make given V as the true original value
$$f(V) = 
     \begin{cases}
       2V - B & B \geq V\\
       0 & B< V
     \end{cases}$$
Where I get confused is when I need to start applying integrals to calculate how to maximize the expected value. 
Thanks for any help.
EDIT: Here is the solution from the book I'm working off of. I do not understand how they are doing the calculus.

Let B be your bid. Let S be the true value of the item. The density function of S equals unity for $0 \leq S \leq 1$, and 0 otherwise.
Your payoff P is 
$$P(S) =
\begin{cases}
  2S - B & B \geq S\\
  0 & \text{otherwise}
\end{cases}$$
The maximum post bid item value is 2, so you should be no more than 2. You want to maximize $E[P(S)]$ with respect to choice of B in the interval [0, 2]. Your expected payoff is:
$$\begin{aligned}
E[P(S)] &=  \int_{S=0}^{S=1} P(S)*1*\,\mathrm{d}S \\
 &=  \int_{S=0}^{S=\min(B,1)} (2S-B)\,\mathrm{d} \\
 &= \left.(S^2-BS)\right|_{S=0}^{S=\min(B,1)} \\
 &= 
     \begin{cases}
       0 &  B\leq1\\
       1 - B &  B>1
     \end{cases}
\end{aligned}$$
so you should bid less than or equal to 1 and expect to break even. 

 A: As I (and others) noted in the comments, the exercise seems to be really confusingly phrased (and not at all like any real auction or other transaction I've ever heard of), and the given solution doesn't look much better.  So instead of trying to explain the book solution, let me rephrase the exercise in a (hopefully) slightly less confusing manner, and then show how I would solve it.

Exercise: You are bidding on an item that has an unknown nominal value $V$ uniformly distributed between $0$ and $1$.  You do not know the nominal value of the item, but you do know that the value of the item to you is twice its nominal value.  You know that, if you bid more than the nominal value of the item, you will win the item and have to pay your bid; otherwise you don't get the item and don't have to pay anything.  How much should you bid to maximize your expected gain?

Solution:
Let $V \sim U(0,1)$ be a random variable denoting the nominal price of the item.  The probability density function of $V$ is $$f(V) = \begin{cases} 1 & \text{if }0 < V < 1 \\ 0 & \text{otherwise.} \end{cases}$$
If you bid an amount $B$, your gain will be $$g(B,V) = \begin{cases} 2V -B & \text{if }V < B \\ 0 & \text{otherwise.} \end{cases}$$
Thus, your expected gain from bidding $B$ is $$\begin{aligned}
\mathbb E_V[g(B,V)] &= \int_{-\infty}^\infty g(B,V)\: f(V)\: \mathrm dV \\
&= \int_0^1 g(B,V)\: \mathrm dV \\
&= \int_0^{\min(1,B)} (2V-B)\: \mathrm dV \\
&= \int_0^{\min(1,B)} 2V\: \mathrm dV - \int_0^{\min(1,B)} B\: \mathrm dV \\
&= V^2 \bigg|_{V=0}^{V=\min(1,B)} - BV \bigg|_{V=0}^{V=\min(1,B)} \\
&= (\min(1,B)^2 - 0^2) - (B\min(1,B) - B\cdot0) \\
&= \min(1,B)^2 - B\min(1,B) \\
&= \begin{cases} 1-B & \text{if }B > 1 \\ 0 & \text{if }B \le 1. \end{cases}
\end{aligned}$$
(Since you said you had trouble following this part in the book solution, I included quite many intermediate steps.  Let me know if there's still something you don't follow.)
As $1-B < 0$ whenever $B > 1$, you should never bid more than $1$.  Instead, any bid of $1$ or less will result in an expected gain of $0$, so any such bid is as good as not bidding at all, which is the optimal strategy.

Actually, this came out looking more like that book solution after all, though hopefully a bit clearer.  What I would probably do, if asked that in an exam or something, would be to start by noting that the chance of winning the item equals $1$ for any bid $B \ge 1$.  Thus, any bid $B > 1$ is clearly suboptimal, as it increases the cost without changing the probability of winning.  That out of the way, I'd the just calculate $$\mathbb E[g(B)]
= \int_0^1 g(B,V)\: \mathrm dV
= \int_0^B (2V-B)\: \mathrm dV
= B^2 - B^2 = 0$$
for all $B \le 1$.  (Or I might note that the integrand $2V-B$ has odd symmetry around the midpoint $V = B/2$ of the integration interval, so the integral has to be zero by symmetry considerations alone.)
A: Edit: This replaces an earlier solution that had a major error. 
We explore the consequences of offering $b$, where $0 \le b\le 1$. Let random variable $W$ denote the value of the item, given that the offer was accepted. For $0 \le w \le b$ we have 
$$Pr(W \le w)=Pr(V\le w|V\le b)=\frac{\Pr(V \le w)}{\Pr(V \le b)}=\frac{w}{b}.$$
Thus $W$ has uniform distribution on $[0,b]$, and therefore has expectation $b/2$.
The profit is $2W-b$. This has expectation $2E(W)-b$, which is $0$ whatever $b$ in $[0,1]$ is chosen for the offer. 
Offering $\lt 0$ trivially also gives expectation of profit equal to $0$. 
If we offer $b\gt 1$, the profit is $2V-b$, which has mean $1-b \lt 0$.
