Question

The area of the region enclosed by the curve \(y=x^3\) and its tangent at the point (−1,−1) is

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

Answer 1

The Correct Option is D

The problem involves finding the area enclosed by the curve \(y = x^3\) and its tangent at the point \((-1, -1)\). We will solve this problem step-by-step.

1. Find the equation of the tangent line at \((-1, -1)\):
   • The derivative of the function \(y = x^3\) gives us the slope of the tangent. The derivative is given by: \(\frac{dy}{dx} = 3x^2\)
   • At the point \((-1, -1)\), the slope is: \(3(-1)^2 = 3\)
   • The equation of the tangent line in point-slope form is: \(y - (-1) = 3(x - (-1))\) Simplifying, the equation becomes: \(y = 3x + 2\)
2. Find the intersection points:
   • To find where this tangent line intersects the curve \(y = x^3\), we set: \(x^3 = 3x + 2\)
   • Rearranging gives: \(x^3 - 3x - 2 = 0\)
   • Testing \(x = -1\) (already a known point) confirms it as a root. Factoring the cubic, we get: \((x + 1)(x^2 - x - 2) = 0\)
   • Further factoring the quadratic \(x^2 - x - 2 = (x - 2)(x + 1)\), the intersections are: \(x = -1, 2\)
3. Calculate the area:
   • The integral for the area between the curve and tangent line from \(-1\) to \(2\) is: \(\int_{-1}^{2} (3x + 2 - x^3) \, dx\)
   • Compute the definite integral:
     • Calculate: \(\int (3x + 2 - x^3) \, dx = \frac{3x^2}{2} + 2x - \frac{x^4}{4}\)
     • Evaluate from \(-1\) to \(2\): \[ \left[\frac{3(2)^2}{2} + 2(2) - \frac{(2)^4}{4}\right] - \left[\frac{3(-1)^2}{2} + 2(-1) - \frac{(-1)^4}{4}\right] \] Simplifying gives: \[ \left[6 + 4 - 4\right] - \left[\frac{3}{2} - 2 - \frac{1}{4}\right] = 6 - \left[\frac{-3}{4}\right] \] \[ = \frac{24}{4} + \frac{3}{4} = \frac{27}{4} \]

Thus, the area of the region enclosed by the curve \(y = x^3\) and its tangent at the point \((-1, -1)\) is \(\frac{27}{4}\), which is the correct answer.