Question

Let A(x) represent the area bounded by the graph and the

horizontal axis and vertical lines at t=0 and t=x for the graph
shown. Evaluate A(x) for x = 1, 2, 3, 4, and 5.

Answer 1

A(1) = 1

A(2) =
`2(1)/(2)`

A(3) =
`4(1)/(2)`

A(4) = 6

A(5) = 7

Explanation:

In this problem, we are asked to solve for the area under the graphed line, bounded by the x-axis and y-axis (which it calls t = 0) for specific right bounds of x = 1, 2, 3, 4, and 5.

For x = 1:

We can see that the horizontal distance (length) of the area under the line is 1, and the vertical distance is a constant 1 as well. Hence, the area is:

`A_\text{rect} = \text{length}\cdot\text{width}`

A(1) = 1 × 1 = 1

For x = 2:

This area is composed of a rectangle of length 2 and height 1 along with a right triangle of base 1 and height 1.

`A_\text{triangle} = (1)/(2)(\text{base}*\text{height})`

`A(2) = A_\text{rect} + A_\text{triangle}`

A(2) = (2 × 1) +
`(1)/(2)(1 * 1)`

A(2) = 2 +
`(1)/(2)`

A(2) =
`2(1)/(2)`

For x = 3:

This area is composed of all of A(2) along with a rectangle of length 1 and width 2.

`A(3) = A(2) + A_\text{rect}`

A(3) =
`2(1)/(2)` + (2 × 1)

A(3) =
`2(1)/(2)` + 2

A(3) =
`4(1)/(2)`

For x = 4:

This area is composed of all of A(3) along with a square of length 1 and a triangle of base 1 and height 1.

`A(4) = A(3) + A_\text{square} + A_\text{triangle}`

`A(4) = A(3) + (1 * 1) + (1)/(2){(1 * 1)`

A(4) =
`4(1)/(2)` + 1 +
`(1)/(2)`

A(4) = 6

For x = 5:

This area is composed of all of A(4) along with a square of length 1.

`A(5) = A(4) + A_\text{square}`

A(5) = 6 + (1 × 1)

A(5) = 6 + 1

A(5) = 7