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

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First we need to calculate equation of the tangent line for which we use the following formula

`y=y_1+f'(x_1)(x-x_1)`

where `y=f(x)` is the curve and `(x_1,y_1)` is the point where the tangent touches the curve.

We will first calculate the derivative `f'(x).`

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

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

Now we calculate the equation of the tangent line.

`y=1+x-(-1)`

`y=x+2`

In order to determine the upper bound of integration we need to calculate the the point of intersection of the tangent line and the graph of the given function. Hence, we need to solve the following system of equations.

`y=x+2`

`y=x^3-2x`

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

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

Factor the equation.

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

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

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

`(x+1)[x(x-1)-2]=0`

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

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

From this we see that the points of intersection are `x=-1` (which we already knew) and `x=2.` Those are also the bounds of integration.

To calculate the area between two curves we simply subtract area under the lower curve from the area under the upper curve. Looking at the image below, we see that the upper curve is the tangent line. Therefore, we get

`A=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=`

`-4+6+4-(-1/4+3/2-2)=6+3/4=27/4`

**The area of the region bounded by the graph of the given function and tangent to the graph at point `(-1,1)` is equal to `27/4.` **