The following holds for definite integrals:
1) $\int_a^b f(x)\, dx =\int_a^c f(x)\,dx+\int_c^b f(x)\,dx\ $, for any numbers $a$, $b$, and $c$
and
2) $\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$.
In your setup, you could use $\int_0^1 f(x)\,dx +\int_1^3f(x)\,dx =\int_0^3f(x)\,dx.$
$7+\int_1^3f(x)\,dx = 4\quad\Rightarrow\quad\int_1^3f(x)\,dx=4-7=-3.$
The definite integral gives an area if the graph of the function you're integrating is above the $x$-axis:
If $f(x)\ge0 $ over $[a,b]$, then $\int_a^bf(x)\,dx$ is the area bounded by the graph of $f$, the $x$-axis, and the lines $x=a$, $x=b$.
Area is nonnegative, but not so for the integral. If the graph of the function you're integrating is below the $x$-axis over $[a,b]$, then $\int_a^b f(x)\,dx$ is negative; but its absolute value is the area bounded by the graph of $f$, the $x$-axis, and the lines $x=a$, $x=b$:
If $f(x)\le0 $ over $[a,b]$, then $\int_a^bf(x)\,dx$ is the negative of the area bounded by the graph of $f$, the $x$-axis, and the lines $x=a$, $x=b$.
In your example, the graph of $f$ would be above the $x$-axis (at least in part) over $[0,1]$ with "area" 7 ($\int_0^1 f(x)\thinspace dx=7$). You were told $\int_0^3 f(x)\,dx=4$. So over $[1,3]$, the graph would have to dip below the $x$-axis: assuming the graph of $f$ is always below the $x$-axis over $[1,3]$, there would have to be 3 units of area below the $x$-axis in order for the integral over $[0,3]$ to be 4, by rule 1) above.
In a nutshell: integrating over an interval over which $f(x)>0$ counts the area as positive, and integrating over an interval over which $f(x)<0$ counts the area as negative.