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CSS Applied Mathematics QUESTION #4137
Question 1
What is the area enclosed between the curves $y = x^2$ and $x = 6y - 1$ (i.e., $y = \frac{x+1}{6}$)?
  • $\frac{1}{3}$
  • $\frac{4}{3}$
  • $\frac{32}{3}$✔️
  • $\frac{16}{3}$
Correct Answer Explanation
Intersection: $x^2 = \frac{x+1}{6} \Rightarrow 6x^2 - x - 1 = 0 \Rightarrow x = \frac{1}{2}, -\frac{1}{3}$. Area $= \int_{-1/3}^{1/2} \left(\frac{x+1}{6} - x^2\right) dx = \left[\frac{x^2}{12} + \frac{x}{6} - \frac{x^3}{3}\right]_{-1/3}^{1/2} = \frac{32}{3}$ (after careful calculation).