Finding Area Bounded by Curve and Lines

We are asked to find the area of the region bounded by the curve \(y^2 = x\), the vertical lines \(x = 1\) and \(x = 4\), and the x-axis.

The equation of the curve is \(y^2 = x\). For \(x \ge 0\), this gives two values for \(y\): \(y = \sqrt{x}\) and \(y = -\sqrt{x}\). Since the region is bounded by the x-axis, we consider the part of the curve above or below the x-axis. The x-axis itself is \(y=0\). The boundaries are \(x=1\) and \(x=4\).

The area of the region bounded by a curve \(y = f(x)\), the x-axis, and the lines \(x=a\) and \(x=b\) is given by the definite integral \(\int_{a}^{b} |f(x)| \, dx\).

In this case, the curve is \(y^2 = x\). For the region above the x-axis, we take the positive square root, so \(y = \sqrt{x}\). The limits of integration are given by the lines \(x=1\) and \(x=4\). Both \(x=1\) and \(x=4\) are within the domain of \(\sqrt{x}\) (\(x \ge 0\)). Also, for \(x\) between 1 and 4, \(\sqrt{x}\) is positive, so \(|f(x)| = f(x)\).

Thus, the area \(A\) is given by the integral:

\(A = \int_{1}^{4} \sqrt{x} \, dx\)

First, let's rewrite \(\sqrt{x}\) using exponents: \(\sqrt{x} = x^{1/2}\).

Now, we integrate \(x^{1/2}\) with respect to \(x\). Using the power rule for integration \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\):

\(\int x^{1/2} \, dx = \frac{x^{1/2 + 1}}{1/2 + 1} = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}\)

Now we evaluate the definite integral using the limits from 1 to 4:

\(A = \left[ \frac{2}{3} x^{3/2} \right]_{1}^{4}\)

Substitute the upper limit (x=4) and the lower limit (x=1) into the expression and subtract the results:

\(A = \left( \frac{2}{3} (4)^{3/2} \right) - \left( \frac{2}{3} (1)^{3/2} \right)\)

Let's calculate the terms separately:

• \(4^{3/2} = (\sqrt{4})^3 = 2^3 = 8\)
• \(1^{3/2} = (\sqrt{1})^3 = 1^3 = 1\)

Substitute these values back into the expression for \(A\):

\(A = \frac{2}{3} (8) - \frac{2}{3} (1)\)

\(A = \frac{16}{3} - \frac{2}{3}\)

Now, subtract the fractions:

\(A = \frac{16 - 2}{3} = \frac{14}{3}\)

The area of the region bounded by the curve \(y^2 = x\), the lines \(x = 1\), \(x = 4\), and the x-axis (considering the part above the x-axis) is \(\frac{14}{3}\) square units. If the question implied the area between the curve and the x-axis (both above and below), we would integrate \(y = \sqrt{x}\) from 1 to 4 and \(y = -\sqrt{x}\) from 1 to 4 and sum their absolute values, but since the curve is symmetric about the x-axis, this would just double the result of integrating \(y = \sqrt{x}\), giving \(2 \times \frac{14}{3} = \frac{28}{3}\). However, the common interpretation for "bounded by the curve and the x-axis" between vertical lines usually implies integrating \(|y|\) or specifying "above the x-axis". Given the options, \(\frac{14}{3}\) is present, suggesting the area above the x-axis was intended.