To solve the problem, we need to understand the integral as the sum of areas under/over the graph of the function.
If the area is above the x-axis, the area is considered positive, otherwise, it's negative.
The function A(x) is defined as:
[tex]A(x)=\int_1^xf(t)dt[/tex]
Find A(2):
[tex]A(2)=\int_1^2f(t)dt[/tex]
The integral from t = 1 to t = 2 is the area of the triangle of base b = 1 unit and height h = 1 unit. Thus the area is:
[tex]A=\frac{bh}{2}=\frac{1\cdot1}{2}=\frac{1}{2}[/tex]
Thus:
A(2) = 1/2
Find A(8):
[tex]A(8)=\int_1^8f(t)dt[/tex]
We have to find the area of the graph from t = 1 to t = 8. Note that the triangle from t =1 to t = 2 has the same dimensions as the triangle from t =2 to t = 3, but the first one is above the x-axis and the second is below, so they cancel out.
We only need to calculate the area from t = 3 to t = 8.
There we have a rectangle of length 8 - 3 = 5 units and a width of -1 unit.
The area of the rectangle is:
A = 5 x (-1) = -5, thus:
A(8) = -5
