Fundamental Theorem of Calculus, integral from upper bound to $x$ Suppose that $f$ is defined on the interval $[a,b]$, where $\;a<b\;$. According to the fundamental theorem of calculus:
$$\frac {d} {dx} \left( \int_{a}^{x}f(t) dt \right) = f(x)$$
What can be said about the following?:
$$\frac {d} {dx} \left( \int_{b}^{x}f(t) dt \right)$$
 A: If you mean $\;x\in[a,b]\;$ and $\;a<b\;$ , then
$$\frac {d} {dx} \left( \int_{b}^{x}f(t) dt\right)=\frac {d} {dx} \left(- \int_{x}^{b}f(t) dt \right)=-\frac {d} {dx} \left( \int_{x}^{b}f(t) dt \right)=-(-f(x))=f(x)$$
A: $$\text{They are equal: }\frac d {dx} \int_a^x f(t)\,dt = \frac d {dx} \int_b^x f(t)\,dt. \tag 1$$
$$ \int_a^x f(t)\,dx = \underbrace{\int_a^b f(t) \,dt}_{\large\text{This is a “constant.''}} + \underbrace{\int_b^x f(t)\,dt.}_{\large\text{This is NOT a “constant.''}} \tag 2$$
“Constant”, in the context of finding $\dfrac d {dx}\Big(\cdots\cdots\Big),$ means not depending on $x$.
When you differentiate both sides of line $(2)$ with respect to $x,$ the constant vanishes, yielding line $(1).$
A: The previous answers are correct, and the value is f(x). From an intuitive point of view, it can be viewed as such: as x changes, the area under the curve between a and x also changes. The fundamental theorem states that the rate of change of that area (meaning the derivative of the integral) is equal to the value of the function f at x. 
Now consider the rate of change of the area between b and x, rather than between a and x. Well, this is just the area between a and x, with some constant term added on (since the area between b and x equals the area between a and x minus the area between a and b). Since that constant is unchanging, it doesn't matter when we consider the rate of change. Thus, the area between b and x changes at the same rate as the area between a and x. That is why the two values you are considering are equal.
This actually applies to any constant value (in the interval on which f is defined) we consider as the lower bound of the integral. 
