I may have misunderstood the proposition, but I thought it was:
Let $f$ be a function $[a,b]\times I\to \Bbb R$, where $I$ is some real interval. Then a sufficient condition for $F(x)=\int^b_af(x,t)\ dt$ to be continuous on $I$ is that it be defined on $I$ and that the restrictions $f_t(x)=f(x,t)$ all be continuous.
But I think there's a counterexample to this that can be constructed using the well-known example of a triangle whose base shrinks while its height increases: the area (integral) stays the same, while the curve itself converges pointwise to $0$.
Define the following on $[0,1]^2$:
$$ f(x, t) = \begin{cases} \frac{2}{x^2}t &\mbox{if } 0 \leq t \leq\frac{x}{2}\\ -\frac{2}{x^2}(t-\frac{x}{2})+\frac{1}{x} & \mbox{if } \frac{x}{2} < t \leq x \\ 0 & \mbox{if } t > x \end{cases} $$
And when $x=0$, set $f(x, t)=0$.
Which should be, for each $x$, an isoceles triangle centered at $\frac{x}{2}$, of height $1/x$ and of base width $x$. All restrictions $f_t$ seem pretty continuous to me, yet the integrals have a whopping discontinuity at $x=0$.