# `f(x) = x^3 , (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. First, determine the slope of the tangent line. Take note that the slope of a line tangent to the curve is equal to the derivative of the function at the point of tangency. So taking the derivative of the function, f'(x) will be:

`f'(x) = d/dx (x^3)`

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

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

The given point of tangency is (1,1). Plugging in x=1 to the derivative yields:

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

`f'(1) = 3`

Hence, the slope of the line tangent to the function at (1,1) is m = 3.

Then, apply the point slope-form to get the equation of the tangent line.

`y - y_1 =m(x - x_1)`

`y - 1 = 3(x - 1)`

Isolating the y, it becomes:

`y - 1 =3x - 3`

`y = 3x - 2`

Thus, the equation of the tangent line is `y = 3x - 2` .

Then, determine the intersection points of  `y = x^3` and  `y = 3x - 2`. To do so, set the two y's equal to each other.

`y = y`

`x^3 = 3x - 2`

Take note that to solve polynomial equation, one side should be zero.

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

Then factor the left side using grouping method.

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

`(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`

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

Set each factor equal to zero.

• `(x-1)^2 = 0`

`x - 1=0`

`x=1`

• `x + 2=0`

`x=-2`

Then, plug-in the x values to either y=x^3 or y = 3x - 2, to get the y coordinates of the intersection.

`x=1`

`y = 1^3=1`

`x=-2`

`y=(-2)^3 = -8`

So, the two equations intersect at (1,1) and (-2,-8). Hence, bounded region of y = x^3 and y = 3x - 2 is:

To determine the area of the bounded region, draw a vertical strip. (See attached image.)

In the figure, the top of the vertical strip touches the graph of y=x^3. And its lower end touches the graph of y = 3x-2. Also, the bounded region starts at x=-2 and ends at x=1.

Applying the formula

`A = int_a^b (y_(_(upper)) - y_(_(lower)))dx`

the integral needed to compute the area of the bounded region is:

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

Evaluating the integral, it results to:

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

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

`A = (1^4/4 - (3*1^2)/2+2*1) - ((-2)^4/4- (3*(-2)^2)/2+2*(-2))`

`A =3/4-(-6)`

`A=27/4`

Therefore, the area of the bounded region is `27/4` square units.

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