# 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 = x^3 - 2x` `(-1, 1)`

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### 1 Answer

First we need to find equation of the tangent line which we can do by using the following formula:

`y=f'(x_0)(x-x_0)+y_0`

In our case `f'(x)=y'=3x^2-2`

`f'(-1)=3(-1)^2-2=1`

By plugging that into the formula for equation of the tangent line we get

`y=1(x+1)+1ž`

`y=x+2`

Now we can draw both graphs of the function and its tangent line (see the image below). However, before we start calculating the area of the region we first need to find point of intersection of the two graphs (point A on the image). We can do that by solving the following system of equations:

`y=x+2`

`y=x^3-2x`

`x+2=x^3-2x`

`x^3-3x-2=0`

`(x-2)(x+1)^2=0`

We already know the two graphs touch at `x=-1` and now we also know they intersect at `x=2.` Therefore to find the area between the two graphs we calculate the area below the tangent line and from it subtract the area below the function:

`int_-1^2((x+2)-(x^3-2x))dx=int_-1^2(-x^3+3x+2)dx=`

`(-x^4/4+(3x^2)/2+2x)|_-1^2=27/4`

**The area is** `27/4.`