# Area in between two curves.

## Homework Statement

Set up and evaluate the definite integral that gives the area of the region bounded by the graph of the function and the tangent line to the graph at the given point

y = f(x) = (x)^3, at (1,1)

## Homework Equations

Area in between two curves

## The Attempt at a Solution

The line tangent to (x)^3 at (1,1) is y = 3x - 2.

So the next thing I want to find are my limits of integration, I tried setting x^3 = 3x - 2 and solving for x which is where I encounter my first problem, when using synthetic division to find the zero's of x^3 - 3x + 2 I get x = -1 and x = 2 which is the opposite of what my little ti-83 shows me. I thought the zeros were of the form ( x - h ) when using synthetic division, no?

When I actually look at the answer it shows the limits of integration as -2 and 1. I can set up the integrand using the fact that y = x^3 is greater than or equal to y = 3x - 2 over [-2,1]; my integrand should be x^3 - 3x + 2.

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hunt_mat
Homework Helper
You started off well, but you had problems when you tried to find in intersection points $x^{3}=3x-2$, trying x=1 shows that 1=3-2=1, so the point x=1 is a solution which is geometrically obvious as that is where the tangent line touches the curve. So now I would write:
$$x^{3}-3x+2=(x-1)(x^{2}+x-2)=(x-1)(x+2)(x-1)$$
So it appears that the two limits are -2 and 1.
The area between the two curves is:
$$\int_{-2}^{1}x^{3}dx-\int_{-2}^{1}3x-2dx$$

got it, thanks.