Answer:
24 units²
Step-by-step explanation:
Given:
To find:
The area enclosed by these lines can be expressed as:
[tex]\displaystyle \large{\int_0^6 \int_2^6 dy dx }[/tex]
For:
[tex]\displaystyle \large{\int_2^6 dy}[/tex]
Recall:
[tex]\displaystyle \large{\int_a^b dy = [y]_b^a = a-b}[/tex]
Therefore:
[tex]\displaystyle \large{\int_2^6 dy = [y]_2^6 = 6-2 = 4}[/tex]
Now we we have:
[tex]\displaystyle \large{\int_0^6 4dx}[/tex]
Recall:
[tex]\displaystyle \large{\int_a^bk dx = [kx]_a^b = kb-ka}[/tex]
for k = constant
Therefore:
[tex]\displaystyle \large{\int_0^6 4 dx = [4x]_0^6 = 4(6)-4(0) = 24}[/tex]
Hence, the area of region is 24 units²