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  • Using integration, find the area of the region enclosed by the parabola  and the line <img alt="3x - y + 6 = 0." src="https://learn.careers360.com/latex

Using integration, find the area of the region enclosed by the parabola y = 3x^{2} and the line 3x - y + 6 = 0.

 

Answers (1)

 y = 3x^{2} and the line 3x - y + 6 = 0.

Required Area = Area of AOB

Point A (-1,3)

Point B(2,12)

\\$Area$ = \int _{-1}^{2}(3x+6 - \sqrt{3}x)dx \\\\ = [\frac{3}{2}x^2+6x - \frac{\sqrt{3}}{2}x^2]_{-1}^2 \\\\ = (\frac{3}{2} - \frac{\sqrt{3}}{2})(4-1) + 6(2+1) \\\\ = \frac{3}{2}(3-\sqrt{3})+18 \ sq \ unit.

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