Key Concepts and Formulas

• Area Between Curves: The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$, where $f(x) \ge g(x)$ on $[a,b]$, is given by $A = \int_a^b [f(x) - g(x)] \, dx$.
• Equation of a Tangent Line: The equation of the tangent line to a curve $y = f(x)$ at a point $(x_0, y_0)$ is $y - y_0 = f'(x_0)(x - x_0)$.
• Derivatives: The power rule for differentiation: if $y = x^n$, then $\frac{dy}{dx} = nx^{n-1}$.

Step-by-Step Solution

Step 1: Find the Equation of the Tangent Line

We are given the curve $y = x^3$ and the point $(-1, -1)$. Our goal is to find the equation of the tangent line to the curve at this point.

1. Find the derivative of the curve: The derivative $\frac{dy}{dx}$ gives the slope of the tangent line at any point $(x, y)$ on the curve. Given: $y = x^3$ Differentiating with respect to $x$: $\frac{dy}{dx} = 3x^2$ Explanation: Applying the power rule to find the derivative of $x^3$.

2. Calculate the slope at the given point: Substitute the x-coordinate of the point of tangency, $x=-1$, into the derivative to find the specific slope $m$ of the tangent line at that point. $m = \left(\frac{dy}{dx}\right)_{x=-1} = 3(-1)^2 = 3(1) = 3$ Explanation: Substituting $x=-1$ into the derivative to find the slope of the tangent line at $x=-1$.

3. Use the point-slope form to find the tangent equation: The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$. Here, $(x_1, y_1) = (-1, -1)$ and $m=3$. Substitute these values into the point-slope form: $y - (-1) = 3(x - (-1))$ $y + 1 = 3(x + 1)$ $y + 1 = 3x + 3$ $y = 3x + 2$ Let's denote this as $y_L = 3x+2$ (the tangent line equation) and the curve as $y_C = x^3$. Explanation: Using the point-slope form of a line to find the equation of the tangent line.

Step 2: Find the Points of Intersection Between the Curve and the Tangent

To find the points where the tangent line intersects the curve $y=x^3$, we set their y-values equal to each other. These intersection points will define the limits of integration.

1. Set the equations equal: Set $y_C = y_L$: $x^3 = 3x + 2$

2. Rearrange into a cubic equation: Rearrange this into a standard cubic equation: $x^3 - 3x - 2 = 0$

3. Solve the cubic equation: Since the line is tangent to the curve at $x=-1$, we know that $x=-1$ is a repeated root. Thus $(x+1)^2$ is a factor of the cubic.

   $x^3 - 3x - 2 = (x+1)(x^2-x-2) = (x+1)(x+1)(x-2) = (x+1)^2(x-2) = 0$ The roots are $x=-1$ and $x=2$.

4. Identify the intersection points: The x-coordinates, $x=-1$ and $x=2$, are the limits of integration for our area calculation.

• For $x=-1$, $y = (-1)^3 = -1$. Point: $(-1, -1)$.
• For $x=2$, $y = (2)^3 = 8$. Point: $(2, 8)$.

Step 3: Set Up the Definite Integral for the Area

We need to find the area between the curve $y_C=x^3$ and the tangent line $y_L=3x+2$ from $x=-1$ to $x=2$.

1. Determine the interval: The intersection points define the interval of integration as $[-1, 2]$.

2. Identify the "upper" and "lower" functions: We need to know which function has a greater y-value in the interval $(-1, 2)$. A simple way to do this is to pick a test point within the open interval, say $x=0$.

   • For the curve $y_C=x^3$: $y_C(0) = 0^3 = 0$.
   • For the tangent $y_L=3x+2$: $y_L(0) = 3(0) + 2 = 2$. Since $y_L(0) = 2 > y_C(0) = 0$, the tangent line $y_L=3x+2$ is above the curve $y_C=x^3$ in the interval $[-1, 2]$.

3. Formulate the integral: The area is given by the integral of (upper function - lower function) with respect to $x$ over the interval. $A = \int_{-1}^{2} (y_L - y_C) \, dx$ $A = \int_{-1}^{2} ((3x + 2) - x^3) \, dx$ $A = \int_{-1}^{2} (3x + 2 - x^3) \, dx$

Step 4: Evaluate the Definite Integral

Now, we evaluate the integral using the power rule for integration and the Fundamental Theorem of Calculus.

1. Integrate each term: The antiderivative of $3x + 2 - x^3$ is: $\int (3x + 2 - x^3) \, dx = \frac{3x^{2}}{2} + 2x - \frac{x^{4}}{4} + C$

2. Apply the limits of integration: $A = \left[ \frac{3}{2}x^2 + 2x - \frac{1}{4}x^4 \right]_{-1}^{2}$ This means we evaluate the antiderivative at the upper limit ($x=2$) and subtract its value at the lower limit ($x=-1$). $A = \left( \frac{3}{2}(2)^2 + 2(2) - \frac{1}{4}(2)^4 \right) - \left( \frac{3}{2}(-1)^2 + 2(-1) - \frac{1}{4}(-1)^4 \right)$

3. Perform the arithmetic carefully:

   First part (evaluate at $x=2$): $\left( \frac{3}{2}(4) + 4 - \frac{1}{4}(16) \right) = 6 + 4 - 4 = 6$

   Second part (evaluate at $x=-1$): $\left( \frac{3}{2}(1) - 2 - \frac{1}{4}(1) \right) = \frac{3}{2} - 2 - \frac{1}{4} = \frac{6}{4} - \frac{8}{4} - \frac{1}{4} = \frac{-3}{4}$

   Subtract the second part from the first part: $A = 6 - \left( -\frac{3}{4} \right) = 6 + \frac{3}{4} = \frac{24}{4} + \frac{3}{4} = \frac{27}{4}$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs when substituting limits of integration.
• Incorrectly Identifying Upper/Lower Functions: Always check which function is greater in the interval of integration.
• Missing Repeated Roots: When a line is tangent to a curve, the point of tangency represents a repeated root when solving for intersection points.

Summary

The area of the region enclosed by the curve $y=x^3$ and its tangent at $(-1,-1)$ is found by integrating the difference between the tangent line and the curve between their intersection points. The correct setup and evaluation of the definite integral leads to the area of $\frac{27}{4}$ square units.

Final Answer The final answer is \boxed{\frac{27}{4}}, which corresponds to option (C).