Question

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 :

• A. \(\frac{110}{9}\)
• B. \(\frac{252}{9}\)
• C. \(\frac{114}{9}\)
• D. \(\frac{126}{9}\)

Answer 1

The Correct Option is D

To find the area of the region bounded by the curve \(y=\sqrt{3x+10}\), the x-axis, and the lines \(x = -3\) and \(x = 2\), we need to set up and evaluate the definite integral of the function from \(-3\) to \(2\). The area \(A\) is given by the integral:

\[ A = \int_{-3}^{2} \sqrt{3x+10} \, dx \]

First, we perform a substitution to simplify the integral. Let \(u = 3x + 10\), then \(du = 3 \, dx\) or \(dx = \frac{1}{3}du\). The limits of integration change with this substitution: when \(x = -3\), \(u = 3(-3) + 10 = 1\) and when \(x = 2\), \(u = 3(2) + 10 = 16\). The integral becomes:

\[ A = \int_{1}^{16} \sqrt{u} \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{16} u^{1/2} \, du \]

The antiderivative of \(u^{1/2}\) is \(\frac{2}{3} u^{3/2}\), so we have:

\[ A = \frac{1}{3} \left[\frac{2}{3} u^{3/2} \right]_{1}^{16} \]

Calculating the definite integral:

\[ A = \frac{1}{3} \left[\frac{2}{3} (16^{3/2}) - \frac{2}{3} (1^{3/2})\right] \]

\[ = \frac{1}{3} \left[\frac{2}{3} \cdot 64 - \frac{2}{3} \cdot 1\right] \]

\[ = \frac{1}{3} \left[\frac{128}{3} - \frac{2}{3}\right] \]

\[ = \frac{1}{3} \times \frac{126}{3} \]

\[ = \frac{126}{9} \]

Thus, the area of the region is \(\frac{126}{9}\).