Question

Given `f(x) = 22x+8`. Then let the area bounded by `f(x): x in (0,x)` and the x-axis be denoted by `A(x)`. Show that `A'(x) = f(x)` ?

Answer 1

`A(x) = 11x^2 + 8x`
`A'(x) = 22x+8 = f(x)`

Explanation:

The area under `f(x) = 22x+8` and the x-axis `in (0,x)`
consists of a triangle of base `x` and height `f(x)-8` plus a rectangle of sides `8` and `x`.

Area of triangle = `1/2 xx Base xx Height`

Area of rectangle = `Base xx Height`

Therefore in this case:
Area of triangle `= 1/2x(22x+8-8)` = `11x^2`
Area of rectangle `=8x`

If the required area is denoted by `A(x)`
`A(x) = 11x^2+8x`

Also `A'(x) = d/dx(11x^2+8x) = 22x+8 = f(x)`

Therefore we have confirmed that `A'(x) = f(x)`