The area of the region bounded by the curve \(y = \sqrt{3x+10}\), x-axis and between the lines x = −3 and x = 2 is equal to:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{126}{9}\)
On \([-3, 2]\), \(3x + 10\) runs from 1 to 16, so the curve lies above the x-axis and the area under the curve is
\( \displaystyle \text{Area} = \int_{-3}^{2}\sqrt{3x+10}\,dx \)
Since \(\int (ax+b)^{1/2}\,dx = \frac{(ax+b)^{3/2}}{\frac{3}{2}a}\), here with a = 3:
\( \displaystyle \text{Area} = \left[\frac{2}{9}(3x+10)^{3/2}\right]_{-3}^{2} \)
\( \displaystyle = \frac{2}{9}\left(16^{3/2} - 1^{3/2}\right) \)
\( \displaystyle = \frac{2}{9}(64 - 1) = \frac{126}{9} \)
(This equals 14 square units.)