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CSS Applied Mathematics QUESTION #1295
Question 1
The area of the region enclosed between \(y=x^2\) and \(y=x+6\) is:
  • 20\(\dfrac{1}{6}\)
  • 14\(\dfrac{2}{3}\)
  • \(\dfrac{125}{6}\)✔️
  • 18
Correct Answer Explanation
Intersections: \(x^2=x+6\Rightarrow x^2-x-6=0\Rightarrow(x-3)(x+2)=0\), so \(x=-2,3\). Area \(=\int_{-2}^{3}(x+6-x^2)\,dx=\left[\dfrac{x^2}{2}+6x-\dfrac{x^3}{3}\right]_{-2}^{3}=\left(\dfrac{9}{2}+18-9\right)-\left(2-12+\dfrac{8}{3}\right)=\dfrac{27}{2}+\dfrac{34}{3}=\dfrac{81+68}{6}=\dfrac{125}{6}\).