Question

The area of the region enclosed by the curve y = x³ and its tangent at the point (-1, -1) is

• A. \(\frac{19}{4}\)
• B. \(\frac{23}{4}\)
• C. \(\frac{27}{4}\)
• D. \(\frac{31}{4}\)

Answer 1

The Correct Option is C

To find the area of the region enclosed by the curve \( y = x^3 \) and its tangent at the point \((-1, -1)\), we follow these steps:

1. First, we determine the equation of the tangent line to the curve at the given point \((-1, -1)\).
2. The derivative of the function \( y = x^3 \) gives the slope of the tangent.

Let's calculate the derivative:

The derivative of \( y = x^3 \) with respect to \( x \) is given by:

\(\frac{dy}{dx} = 3x^2\)

Now, substituting \( x = -1 \) into the derivative to find the slope of the tangent at \((-1, -1)\):

3(-1)^2 = 3 \times 1 = 3\)

This means the slope of the tangent line at \((-1, -1)\) is 3.

Using the slope-point form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \((x_1, y_1) = (-1, -1)\) and \( m = 3 \), the equation of the tangent line is:

y + 1 = 3(x + 1)

Simplifying gives us:

y = 3x + 3 - 1

y = 3x + 2

Next, we find the points of intersection between the curve \( y = x^3 \) and the tangent line \( y = 3x + 2 \).

3. Set the equation of the curve equal to the equation of the tangent:

x^3 = 3x + 2

Rearranging gives us the equation:

x^3 - 3x - 2 = 0

This cubic equation can be solved by trial and error or using the rational root theorem:

• x = -1 is a root since substituting it gives zero: \((-1)^3 - 3(-1) - 2 = -1 + 3 - 2 = 0\).

Factoring the cubic using \( x = -1 \) gives:

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

The quadratic \( x^2 - x - 2 \) can be factored further as:

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

Thus, the roots are \( x = -1, x = 2 \).

4. The area \( A \) between the curves from \( x = -1 \) to \( x = 2 \) is given by the integral:

A = \int_{-1}^{2} [ (3x + 2) - (x^3) ] \,dx

Let's evaluate this integral:

A = \int_{-1}^{2} (3x + 2 - x^3) \,dx

= \left[ \frac{3x^2}{2} + 2x - \frac{x^4}{4} \right]_{-1}^{2}

Compute the definite integral:

• Substitute \( x = 2 \):
• \left[ \frac{3(2)^2}{2} + 2(2) - \frac{(2)^4}{4} \right]\)
• = \left[ \frac{12}{2} + 4 - \frac{16}{4} \right]
• = \left[ 6 + 4 - 4 \right]
• = 6
• Substitute \( x = -1 \):
• \left[ \frac{3(-1)^2}{2} + 2(-1) - \frac{(-1)^4}{4} \right]\)
• = \left[ \frac{3}{2} - 2 - \frac{1}{4} \right]
• = \left[ \frac{3}{2} - \frac{8}{4} - \frac{1}{4} \right]
• = \left[ \frac{3}{2} - \frac{9}{4} \right]
• = \left[ \frac{6}{4} - \frac{9}{4} \right]
• = \left[ -\frac{3}{4} \right]

Thus, the area is:

A = 6 + \frac{3}{4} = \frac{27}{4}

Therefore, the correct answer is \(\frac{27}{4}\).