What is the probability $X + Y > 100$? $X,Y \sim U[40,80]$ I'm struggling to write out the integral for this. Both $x$ and $y$ are i.i.d. distributed uniformly over $[40,80]$.
I think it should be of the form $\int_{40}^{80} \int_{100-x}^{80} \frac{1}{40\times40}dy~dx$ but this evaluates to $1$?
Thanks for your help!
 A: The trick to these questions is to draw a picture.  (In fact, I'd contend one should always attempt to draw a representative picture for math problems.  The process of thinking about how to draw the picture is helpful to solve the problem.)
We're concerned with the region $x+y > 100$, where $x$ and $y$ are uniformly distributed from $40$ to $80$.  So, let's plot $x+y=100$ and some bounds for $x=40$, $x=80$, $y=40$, and $y=80$:

Now, we return to your integral:
\begin{align}
P[x+y \ge 100\; |\; x,y \in A] &= \iint_A f_{X,Y}(x,y) \; dA \\
&= \iint_A f_X(x)\cdot f_Y(y) \; dA \quad \text{(because $X$ and $Y$ are iid)} \\
&= \iint_A \frac{1}{40}\cdot\frac{1}{40} \; dA \\
&= \frac{1}{1600} \iint_A \; dA
\end{align}
Now, this raises the question --- how do we compute that integral?  Fortunately, we don't have to!  That integral represents the shaded area in the above graph.  However, even that is kind-of nasty to find.  Thus, we can use our complement rule and realize that the shaded area is just $1-(\text{probability represented by the missing corner})$.  
That missing corner is just a triangle of base and height equal to $20$, so its area is $200$ square units.  Let's denote that triangular region by $A^c$ (for $A$'s complement).  Thus: 
\begin{align}
P[x+y \ge 100\; |\; x,y \in A] &= \frac{1}{1600} \iint_A \; dA \\
&= 1 - \left(\frac{1}{1600} \iint_{A^c} \; dA^c \right) \\
&= 1 - \left(\frac{1}{1600} \cdot 200 \right) \\
&= 1 - \frac{1}{8} \\
&= \frac{7}{8}
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{40}^{80}{1 \over 40}
\int_{40}^{80}{1 \over 40}\bracks{x + y > 100}
\dd x\,\dd y}
\\[5mm] = &\
{1 \over 1600}\int_{40}^{80}\int_{40}^{80}
\bracks{x > 100 - y}\dd x\,\dd y
\\[1cm] = &\
{1 \over 1600}\ \times
\\[2mm] &\
\int_{40}^{80}\!\!\braces{\!\!%
\bracks{100 - y < 40}\int_{40}^{80}\dd x + 
\bracks{40 < 100 - y < 80}\int_{100 - y}^{80}\dd x\!\!}\!\dd y
\\[1cm] = &\
{1 \over 1600}\braces{%
\int_{40}^{80}\bracks{y > 60}40\,\dd y
+ \int_{40}^{80}
\bracks{y < 60}\pars{y - 20}\,\dd y
}
\\[5mm] = &\
{1 \over 40}\int_{60}^{80}\dd y +
{1 \over 1600}\int_{40}^{60}\pars{y - 20}\dd y =
\bbx{7 \over 8} = 0.875
\end{align}
A: Comment. I used a simulation in R to make the picture suggested by @EthanBolker.
set.seed(1214);  m = 30000
x = runif(m, 40, 80);  y = runif(m, 40, 80)
s = x + y;  cond = (s > 100)
plot(x, y, pch=".")
points(x[cond], y[cond], pch=".", col="green2")
mean(cond)
[1] 0.8707


The distribution over the square is uniform. So it is enough to figure out
what fraction of the square is colored green. 
To make a nice picture $m = 30,000$
points is about right. But for a really good approximation to the right answer,
not nearly enough. Even so, your answer ought to be somewhere near 0.87 (as elegantly shown by @apnorton).
Although it is not necessary to do so, it might be good practice to set up the
limits for the double integral over the black region and subtract from $1.$ What is the (constant) value of the joint
density function $f(x,y)$ over the square with vertices at $(40,40)$ and $(80,80)?$ Obviously, that's not completely trivial, because we
accumulated one wrong answer on this page (now disappeared) while I was fussing with my simulation (and
answering a couple of telemarketing calls). 
