In[1]:=
Int1 = Integrate[(-2 t)*4, {t,0,1}]
Out[1]=
-4
On the second leg, we can take x=4 - 2 t, y = 2, 0 <= t <= 1:
In[2]:=
Int2 = Integrate[(-2)*(-2), {t,0,1}]
Out[2]=
4
On the third leg, we can take x=2 t + 2, y = 2+ 4 t, 0 <= t <= 1:
In[3]:=
Int3 = Integrate[(-2 - 4 t)*(2), {t,0,1}]
Out[3]=
-8
On the fourth leg, we return to the origin and can take x=4 - 4 t, y = 6 - 6 t, 0 <= t <= 1:
In[4]:=
Int4 = Integrate[(-6 + 6 t)*(-4), {t,0,1}]
Out[4]=
12
In[5]:=
Area = Int1 + Int2 + Int3 + Int4
Out[5]=
4
The explicit expression asked for is
Area = Integrate[(-2 t)*4, {t,0,1}]
+ Integrate[(-2)*(-2), {t,0,1}]
+ Integrate[(-2 - 4 t)*(2), {t,0,1}]
+Integrate[(-6 + 6 t)*(-4), {t,0,1}],
and its value is 4. You could equally well calculate the integral of x dy.
Up to Solutions to Test 4