Question

How to solve this problem? I have no Idea, how can I solve it

Answer 1

Please see below.

Explanation:

`F(a) = int_a^a f(t) dt = 0`, and also

`F(a) = a^3-2a^2+a-a`

Setting `a^3-2a^2+a-a = 0` and solving for `a != 0`, we get `a=2`

Now we have
`F(x) = x^3-2x^2+x-2 = x^2(x-2)+1(x-2) = (x^2+1)(x-2)`, so

`F(x) > 0` for `x > 2`.

`f(x) = F'(x) = 3x^2-4x+1`, so the graph of `f` is a parabola that opens upward.

The enclosed area is below the `x`-axis and between the `x`-intercepts.

The `x`-intercepts are `1/3` and `1`, so the required area is:

`int_(1/3)^1 (0-f(x)) dx = -[x^3-2x^2+x]_(1/3)^1 = 4/27`