Answer:
[tex]A(4)=-4\pi [/tex]
[tex]A(8)=-4\pi +8[/tex]
[tex]A(12)=-4\pi +16[/tex]
[tex]A(14)=-4\pi +15[/tex]
Step-by-step explanation:
we are given
[tex]A(x)=\int\limits^x_0 f{x} \, dx[/tex]
Calculation of A(4):
we can plug x=4
[tex]A(4)=\int\limits^4_0 f{x} \, dx[/tex]
Since, this curve is below x-axis
so, the value of integral must be negative
and it is quarter of circle
so, we can find area of quarter circle
radius =4
[tex]A(4)=-\frac{1}{4}\times \pi \times (4)^2[/tex]
[tex]A(4)=-4\pi [/tex]
Calculation of A(8):
we can plug x=8
[tex]A(8)=\int\limits^8_0 f{x} \, dx[/tex]
we can break into two parts
[tex]A(8)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx[/tex]
now, we can find area and then combine them
[tex]A(8)=-4\pi +\frac{1}{2}\times 4\times 4[/tex]
[tex]A(8)=-4\pi +8[/tex]
Calculation of A(12):
we can plug x=12
[tex]A(12)=\int\limits^12_0 f{x} \, dx[/tex]
we can break into two parts
[tex]A(12)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx[/tex]
now, we can find area and then combine them
[tex]A(12)=-4\pi +\frac{1}{2}\times 8\times 4[/tex]
[tex]A(12)=-4\pi +16[/tex]
Calculation of A(14):
we can plug x=14
[tex]A(14)=\int\limits^14_0 f{x} \, dx[/tex]
we can break into two parts
[tex]A(14)=\int\limits^4_0 f{x} \, dx+\int\limits^8_4 f{x} \, dx+\int\limits^12_8 f{x} \, dx+\int\limits^14_12 f{x} \, dx[/tex]
now, we can find area and then combine them
[tex]A(14)=-4\pi +\frac{1}{2}\times 8\times 4-\frac{1}{2}\times 1\times 2[/tex]
[tex]A(14)=-4\pi +16-1[/tex]
[tex]A(14)=-4\pi +15[/tex]
