verify that the solution $u''=f(x)$, $u(0)=u(1)=0$ is given by $u(x)=\int_0^1k(x,y)f(y)dy$ 
verify that the solution $u''=f(x)$, $u(0)=u(1)=0$ is given by $u(x)=\int_0^1k(x,y)f(y)dy$ where 
$k(x,y) =
\begin{cases}
y(x-1),  & \text{ $0\leq y<x\leq 1$} \\[2ex]
x(y-1), & \text{ $0 \leq x<y\leq 1$}
\end{cases}$

EDIT according to feedback
Solution: 
Suppose $0\leq a \leq 1$ and assume f(x) is continuous
\begin{align}
u(x) &= \int_0^a y(x-1)f(y)dy+\int_a^1 x(y-1)f(y)dy\\
u'(x) &= \frac{d}{dx}(\int_0^a y(x-1)f(y)dy+\int_a^1 x(y-1)f(y)dy)\\
& = \int_0^ayf(y)dy+\int_a^1(y-1)f(y)dy\\
u'' &= 0
\end{align}
which is not the solution given. Where did I go wrong?
 A: To split the integrand into two continuous function make $a$ a function of $x$:
\begin{align*}
  a(x) = x.
\end{align*}
Then
\begin{align*}
   \int_0^1 k(x,y) f(y) dy &= \int_0^x y(x-1) f(y) \ dy + \int_x^1 x(y-1) f(y)\ dy.
\end{align*}
To evaluate derivatives of integrals with variable limits use the Leibniz integral rule. The first and second terms are
\begin{align*}
   \frac{d}{dx} \int_0^x y(x-1) f(y) dy &= \int_0^x yf(y) \ dy + x(x-1)\\
   \frac{d}{dx} \int_x^1 x(y-1) dy &= \int_x^1(y-1)f(y) \ dy - x(x-1).
\end{align*}
Summing these and using linearity to collect terms $yf(y)$,
\begin{align*}
   u'(x) = \int_0^1 y f(y) \ dy - \int_x^1 f(y) dy.
\end{align*}
The first term is a constant and the second term can be differentiated by the Fundamental Theorem of Calculus to get
\begin{align*}
   u''(x) = f(x).
\end{align*}
And the boundary conditions are satisfied because $k(0,y)=k(1,y)=0$ for $y\in(0,1)$.
A: Since $u$ doesn't appear on the RHS of the differential equation, it's easy to solve just by taking primitives (assuming as you say that $f$ is continuous). The following is quite routine, actually.
Start with the ODE $$\begin{align}  u''(x) &= f(x) \implies \\ \implies u'(x) &= \int_0^x f(y)\,dy + A \\ \implies u(x) &= \int_0^x\int_0^z f(y)\,dy\,dz + Ax + B\end{align}$$
for arbitrary constants $A,B\in\Bbb R$. We can determine these by the boundary conditions: $$u(0) = \int_0^0\int_0^z f(y)\,dy\,dz + A\cdot 0 + B = B =0$$
and $$u(1) = \int_0^1\int_0^z f(y)\,dy\,dz + A = 0 \implies \\ \implies A = -\int_0^1\int_0^z f(y)\,dy\,dz $$
Now we just write the result, and manipulate it to look like $\int_0^1k(x,y)f(y)dy$ for some function $k$, which should turn out to be yours: $$\begin{align}u(x) &= \int_0^x\int_0^z f(y)\,dy\,dz - x\int_0^1\int_0^z f(y)\,dy\,dz \\ &= \int_0^1\chi_{(0,x)}(y)\int_0^z f(y)\,dy\,dz - x\int_0^1\int_0^z f(y)\,dy\,dz \\ &= \int_0^1\int_0^z \chi_{(0,x)}(y)f(y)\,dy\,dz - x\int_0^1\int_0^z f(y)\,dy\,dz \\ &= \int_0^1\int_0^z(\chi_{(0,x)}(y) - x)f(y)\,dy\,dz\end{align}$$
Almost there, we just need to use Fubini to change the order of integration.  The result is: $$u(x) = \int_0^1\int_y^1(\chi_{(0,x)}(y) - x)f(y)\,dz\,dy = \\ = \int_0^1\underbrace{\left(\int_y^1(\chi_{(0,x)}(y) - x)\,dz\right)}_{k(x,y)}\,f(y)\,dy$$
Now just check that $k(x,y)$ is the function you want.
