Hello, gracy!
Make some sketches and use babytalk . . .
$\displaystyle \int^1_0f(x)\,dx\:=\:7$ and $\displaystyle \int^3_0f(x)\,dx\:=\:4$
Find: $\displaystyle \int^3_1f(x)\,dx$
The first integral says: the area under $\displaystyle f(x)$ from 0 to 1 is $\displaystyle 7$ square units.
The graph might look like this. Code:

*
 *
 *
 7 *

+*
 1

The second integral says: the area under $\displaystyle f(x)$ from 0 to 3 is only $\displaystyle 4.$
How did we lose some area?
Some of the area from 1 to 3 must be negative.
. . The graph must dip below the xaxis.
The graph might look like this. Code:

*
 *
 *
 7 *
 3
+*+
 1 * 3 :
 * :
 *
And that is why: .$\displaystyle \int^3_1f(x)\,dx\:=\:3$