Question

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. $$ y=x^{3}-2 x, \quad(-1,1) $$

Short answer

The area bounded by the curve \(y = x^3 - 2x\) and its tangent line at the point \(-1,1\) is 3.5 square units.

Step-by-step solution

Find the equation of the tangent line

Firstly, derive the function \(y = x^3 - 2x\). The derivative of this function is \(y' = 3x^2 - 2\). Then, plug the given point \(-1, 1\) into the derivative to find the slope of the tangent line: \(m = 3(-1)^2 - 2 = 1\). With the slope and a point, the equation of the tangent line can be found using the point-slope formula: \(y - 1 = 1(x + 1)\), which simplifies to \(y = x + 2\).

Set up the integral

To find the area between the function and its tangent line, we need to set up an integral of the difference between the function and the tangent line. We are going to find this area in the intersecting region, hence, we first need to find the points of intersection between the function and the tangent line by setting the two functions equal to each other: \(x^3 - 2x = x + 2\). This system simplifies to \(x^3 - 3x - 2 = 0\), whose solutions are \(x = -1, 1, 2\). As the tangent point is -1, we integrate the difference of the functions from -1 to the other intersection point, which is 2: \(\int_{-1}^{2} [(x^3 - 2x) - (x + 2)] dx\).

Evaluate the integral

Evaluate the definite integral: \(\int_{-1}^{2} (x^3 - 3x - 2) dx\). To do this, find the antiderivative of the function \(x^3 - 3x - 2\), which is \((1/4)x^4 - (3/2)x^2 - 2x\), and apply the Fundamental Theorem of Calculus by subtracting the antiderivative evaluated at -1 from that evaluated at 2. This results in \(((1/4)2^4 - (3/2)2^2 - 4) - ((1/4)(-1)^4 - (3/2)(-1)^2 + 2) = -3.5\). The negative value signifies that the function is below the tangent line. But the area itself is not negative. Hence, we take the absolute value to find the area, which equals 3.5.