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NDA Math Area Under the Curves

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For the next three (03) items that follow:

The line

2y=3x+12

cuts the parabola

4y=3x^2

[NDA II 2014]

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Question : 24 of 27

Marks: +1, -0

What is the area enclosed by the parabola, the line and the y-axis in the first quadrant?

7 square unit
14 square unit
20 square unit
21 square unit

Solution:

Area

=\left|\int\limits_{-3}^{4}\left(\frac{3}{4}x^{2}-\frac{3}{2}x-6\right)dx\right|

\left|\left[ \frac{3}{4}\frac{x^{3}}{3} - \frac{3}{2}\frac{x^{2}}{2} - 6x \right]_{-2}^{4} = \left[ \frac{4^{3}}{4} - \frac{3}{2}\frac{4^{2}}{2} - 6 \times 4 \right] - \left[ \frac{(0)^{3}}{4} - \frac{3}{2} \times \frac{0^{2}}{2} - 6(0) \right]\right|

=\left|16-12-24+0\right|=\left|-20\right|=20

square units

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