Problem 31 - Calculus Early Transcendentals

Question

Find the area of the following regions by (a) integrating with respect to \(x,\) and (b) integrating with respect to \(y\) Be sure your results agree. Sketch the bounding curves and the region in question. The region bounded by \(y=2 x-1\) and \(x=y^{2}\)

Step-by-Step Solution

Answer

#tag_title# Short Answer #tag_content# The area of the region bounded by the functions \(y = 2x - 1\) and \(x = y^2\) is \(\frac{4}{3}\) square units. The intersection points of the two functions are \((1, -1)\) and \((1, 1)\), and when graphed, the region of interest lies between the line and the parabola.

Step 1: STEP 1

First, let's find the intersection points of the two functions. We have \(y = 2x - 1\) and \(x = y^2\). To find the points where these functions intersect, we can set them equal to each other and solve for one of the variables. When setting \(y\) equal to each other, we get: \(2x - 1 = y^2\) Now, substitute \(x = y^2\) into the equation: \(2(y^2) - 1 = y^2\) Solve for \(y\): \(y^2 = 1\) \(y = \pm 1\) For each value of \(y\), find the corresponding \(x\) values using \(x=y^2\): When \(y = 1\), \(x = 1^2 = 1\) When \(y = -1\), \(x = (-1)^2 = 1\) So the intersection points are \((1, -1)\) and \((1, 1)\).

Step 2: STEP 2

Next, we need to determine which function is higher in the region of interest. Since both points of intersection have the same \(x\)-coordinate (\(x = 1\)), we can distinguish which function is higher by comparing them at any point within the region. Let's take a point between \(y = -1\) and \(y = 1\), say \(y = 0\). \(y = 2x - 1\) evaluated at \(y = 0\) gives \(x = 0.5\). \(y = 0\) happens between \(-1\) and \(1\), so at \(x = 0.5\), \(y=2x-1\) is above \(x=y^2\).

Step 3: STEP 3

Now, let's integrate with respect to x, the difference between the two functions: \(A_x = \int_{-1}^1 (2x - 1 - y^2) dy\) To find the corresponding function in terms of \(x\), we will use \(x = y^2\): \(A_x = \int_{-1}^1 (2y^2 - 1 - y^2) dy\) \(A_x = \int_{-1}^1 (y^2 - 1) dy\) To find the area, integrate: \(A_x = \left[\frac{1}{3}y^3 - y\right]_{-1}^1\) \(A_x = \left(\frac{1}{3}(1)^3 - (1) - \left((-\frac{1}{3}(1)^3 + 1)\right)\right) = \frac{4}{3}\)

Step 4: STEP 4

Now, let's integrate with respect to \(y\). First, we need to find the function in terms of \(y\). We will use \(y = 2x - 1\). \(x = \frac{1}{2}(y+1)\) \(A_y = \int_{1}^{-1} \left(\frac{1}{2}(y+1) - y^2\right) dx\) To find the area, integrate: \(A_y = \left[\frac{1}{4}y^2+y-\frac{1}{3}y^3\right]_1^{-1}\) \(A_y = \left(\frac{1}{4}(-1)^2-(-1)-\frac{1}{3}(-1)^3\right) - \left(\frac{1}{4}(1)^2+(1)-\frac{1}{3}(1)^3\right) = \frac{4}{3}\)

Step 5: STEP 5

Results verification Since both \(A_x\) and \(A_y\) are equal to \(\frac{4}{3}\), the results are consistent, and the area of the region bounded by the given functions is \(\frac{4}{3}\) square units.

Step 6: STEP 6

Finally, let's sketch the bounding curves and the region in question. To sketch the region, plot \(y = 2x - 1\) (a straight line), and \(x = y^2\) (a parabola opening to the right). Mark the intersection points \((1, -1)\) and \((1, 1)\), which correspond to the limits of integration. The area of the region bounded by these functions lies between the line and the parabola, and its value is \(\frac{4}{3}\) square units as calculated above.

Key Concepts

intersection points

To find the area of a region bounded by two curves, it's crucial to first identify their intersection points. These points help to define the limits of integration. In our problem, we're dealing with a line given by the equation \( y = 2x - 1 \) and a parabola represented by \( x = y^2 \). To find where these intersect, equate the equations: \( y = 2x - 1 = y^2 \). By solving for \( y \), we find two solutions: \( y = 1 \) and \( y = -1 \).

Next, we find the corresponding \( x \)-coordinates using \( x = y^2 \):

• At \( y = 1 \), \( x = 1^2 = 1 \).
• At \( y = -1 \), \( x = (-1)^2 = 1 \).

Thus, the intersection points are \((1, 1)\) and \((1, -1)\). These points are crucial as they set the boundaries for our integration process.

bounding curves

Bounding curves define the shape and limits of a region. In this exercise, we have:

• The **line**: \( y = 2x - 1 \). It can be rearranged as \( x = \frac{y+1}{2} \).
• The **parabola**: \( x = y^2 \), opening to the right.

These two functions create a boundary around a 'curved' region. To determine which curve is on top, we compare their values between the intersection points. A common way is to choose a midpoint value; when \( y \) is 0, compute the \( x \)-values: \( x = 0.5 \) from the line, and \( x = 0 \) from the parabola.

Since \( x = 0.5 \) for the line is greater than \( x = 0 \) for the parabola, \( y = 2x - 1 \) lies above **\( x = y^2 \)** between the intersection points.

parabolic region

The region under study is parabolic due to the nature of the equations. Visually, it is an area between a straight line and a curve. This combination creates a unique shape where one boundary arcs and the other stretches linearly.

The sketch of the region can be imagined by plotting \( y = 2x - 1 \) as a straight line sloping upwards and \( x = y^2 \) as a parabola curving to the right. The region of interest is where these two graphs overlap, forming an area that must be integrated to find its size.

area calculation

Calculating the area involves setting up two separate integrals, one with respect to \( x \) and another with respect to \( y \):

• **With respect to \( x \):** Integrate the difference \( (2x - 1) - (y^2) \):
  \[ A_x = int_{-1}^{1} (2x - 1 - x^{2}) \, dy = \left[ \frac{1}{3}y^3 - y \right]_{-1}^{1} \]
• **With respect to \( y \):** Here we use the functions rearranged: \( \frac{1}{2}(y + 1) \) for the line, and \( y^2 \) remains for the parabola:
  \[ A_y = int_{1}^{-1} \left(\frac{1}{2}(y + 1) - y^2\right) \, dx = \left[ \frac{1}{4}y^2 + y - \frac{1}{3}y^3 \right]_{1}^{-1} \]

Both integrals, when evaluated, yield the same area: \( \frac{4}{3} \) square units. Consistency validates that the methods of integration correctly account for the given bounded region.