Question

In Exercises 35 and \(36,\) find the area of the region by subtracting the area of a triangular region from the area of a larger region. The region on or above the $$x$$ -axis bounded by the curves $$y^{2}=x+3$$ and $$y=2 x$$

Short answer

The area of the region on or above the x-axis bounded by the curves \(y^{2}=x+3\) and \(y=2x\) is obtained by subtracting the area below the curve \(y=2x\) from the area under the curve \(y^{2}=x+3\). The exact value of the area will be obtained by computing the definite integral expressions obtained in Step 2.

Step-by-step solution

Identify the intersection points

We need to find the points where \(y^{2} = x + 3\) and \(y = 2x\) intersect. To do this, we substitute \(y = 2x\) into \(y^{2} = x + 3\) and solve for \(x\): \((2x)^{2} = x + 3 \) which simplifies to \(4x^{2} = x + 3\). This is a quadratic equation that can be solved to give \(x = 0, -3\). Replacing \(x\) in \(y = 2x\), we find that the corresponding \(y\) are \(0\) and \(-6\). Hence the curves intersect at points (0,0) and (-3,-6).

Set up the integral for the area

We calculate the area by integrating the absolute value of the difference of the two equations with respect to \(x\), from the lower limit of \(x = -3\) to the upper limit of \(x = 0\). Set up the integral as follows: \( \int_{-3}^{0} [2x - \sqrt{x+3}]\,dx \) and \( \int_{-3}^{0} [2x + \sqrt{x+3}]\,dx\). For \(y = 2x - \sqrt{x+3}\) we only consider when \(2x > \sqrt{x+3}\), and for \(y = 2x + \sqrt{x+3}\), we only consider when \(2x < \sqrt{x+3}\).

Solve the integrals

Evaluate the two integrals separately and sum the resultant areas to obtain the total area. The results should be two numerical values. Then we subtract the smaller from the larger value since the task is to find the area by subtracting the area of a triangular region from the area of a larger region.