The area under a curve between two points can be found out by doing the integral between the two points. In other words, the integral
[tex]\int ^7_0f(x)dx\text{ = Area betw}een\text{ x=1 and x=2 + Area betw}een\text{ x=2 and x=4 + Area betwe}en\text{ x=4 and x=5 - Area betw}een\text{ x=5 and x=7}[/tex]
Let's make a picture of the problem
Then, the integral will be equal to
[tex]\int ^7_0f(x)dx\text{ = Area black zone + Area red zone + Area gr}een\text{ zone - Area blue zone}[/tex]
The area of the black region is given by the area of the triangular part plus the rectangular part, that is
[tex]\begin{gathered} \text{ Area black zone = }\frac{1}{2}2\times1+2\times1 \\ \text{ Area black zone =}1+2 \\ \text{ Area black zone =}3 \end{gathered}[/tex]
The area of the red zone is the area of the rectangle
[tex]\begin{gathered} \text{ Area red zone = 2}\times2 \\ \text{ Area red zone =}4 \end{gathered}[/tex]
The green area is equal to the area of the green triangle,
[tex]\begin{gathered} \text{ Area gre}en\text{ zone=}\frac{1}{2}1\times2 \\ \text{ Area gre}en\text{ zone=}1 \end{gathered}[/tex]
and the blue area is the area of the blue triangle,
[tex]\begin{gathered} \text{ Area blue zone = }\frac{1}{2}2\times2 \\ \text{ Area blue zone = }2 \end{gathered}[/tex]
By substituting these values, the integral is given by
[tex]\int ^7_0f(x)dx\text{ = }3+4+1-2[/tex]
Therefore, the answer is:
[tex]\int ^7_0f(x)dx\text{ = }6[/tex]
