Compute the desired area

Question

I solved it on my own but I need some feedback on whether I solved it correctly or not.

Consider the function $$f(x)=x^3-2x^2-x+2$$

Let $h(x)$ be the function that results when $f(x)$ is shifted by $−1$ in the $x$-direction.

a) Sketch the graph of the function $f(x)$.

I sketched it.

b) Sketch the graph of the function $h(x)$ and give an algebraic definition for $h(x)$.

I sketched it and I formulated the formula $$h(x)=f(x+1)=(x+1)^3-2(x+1)^2-(x+1)+2= x^3-2x^2-2x$$

Is this correct?

c) Find the intersection points of $f(x)$ and $h(x)$.

I solved this simultaneously and got $x=-2$ Then I plugged this value in at $f(x)$ and got $y=-12$

d) Compute the desired area.

this is the part i am confused about. At what interval do I need to do this? -2 to 2? I used -2 to 2 and got the Area= 16.

Answer 1

I think you are supposed to find this area ($f(x)$ is in red, $h(x)$ is in blue):

First, we wish to find the $2$ intersection points. Note that $h(x) = f(x + 1) = x^{3} -x^{2} + 2x$. Then, we want to find:

$$f(x) = h(x)$$

$$x^{3}-2x^{2} - x + 2 = x^{3} + x^{2} -2x$$

$$3x^{2} - x - 2 = 0$$

$$(3x + 2)(x - 1) = 0$$

$$x = -\frac{2}{3}, 1$$

Note that since $f(x)>h(x)$ for $x\in(-\frac{2}{3}, 1)$, we wish to find:

$$\int_{-\frac{2}{3}}^{1}f(x)-h(x)\ dx$$

$$\int_{-\frac{2}{3}}^{1}-3x^{2} + x + 2\ dx$$

$$(-x^{3} + \frac{x^{2}}{2} + 2x)\bigg\vert_{-\frac{2}{3}}^{1}$$

$$\frac{3}{2} + \frac{22}{27}$$

$$\boxed{\frac{125}{54}}$$

Answer 2

You have a mistake in your $h(x)$ and intersection points. Curves intersect at two points.

$h(x) = (x+1)^3 - 2 (x+1)^2 - (x+1) + 2 = x^3 + x^2 - 2x$

Please see the graph for area bound by them. You can use single integral of function $(f(x)-h(x))$ between bounds of $x$ to find area.

Answer 3

[F(x)-H(x)] from -2/3 to 1, where H'=h and F'=f u can see the desired area in graph-picture