Problem 47 - Calculus Early Transcendentals

Question

Use the most efficient strategy for computing the area of the following regions. The region bounded by \(y=\sqrt{x}, y=2 x-15,\) and \(y=0\)

Step-by-Step Solution

Answer

Answer: The area of the region bounded by the given functions is 14 square units.

Step 1: 1. Find Points of Intersection

First, to find where the curves intersect, we must solve for \(x\) in terms of \(y\) for the given functions. Then, we can set them equal to each other to find the intersection points. \(y = \sqrt{x} \Rightarrow x = y^2\) \(y = 2x -15 \Rightarrow x = \frac{1}{2}(y + 15)\) To find the intersection points, set the \(x\) expressions equal to each other: \(y^2 = \frac{1}{2}(y + 15)\) Solving for \(y\): \(2y^2 - y - 15=0\) By factoring or using the quadratic formula, we find the two intersection points: \(y = 3\) and \(y = -5\) Now, find the corresponding \(x\) values: For \(y = 3:\) \(x = 3^2 = 9\) For \(y = -5:\) \(x = \frac{1}{2}(-5 + 15) = 5\) So, the intersection points are \((5, -5)\) and \((9, 3)\).

Step 2: 2. Set up the integral

From the intersection points, we can see that the region can be split into two parts vertically. From \(y=-5\) to \(y=0\), the right boundary is created by \(y=\sqrt{x}\), and for \(y=0\) to \(y=3\), it is created by \(y=2x-15\). In both regions, the left boundary is created by the \(y\)-axis (\(x = 0\)). Thus, the integral formula to find the area is: \(A = \int_{-5}^{0} [\sqrt{x} - 0] dy + \int_{0}^{3} [(2x-15) - 0] dy\) To find the integrals, first convert each integral from \(dy\) to \(dx\): For the first integral, \(y=\sqrt{x} \Rightarrow dy = \frac{1}{2\sqrt{x}} dx\) For the second integral, \(y=2x-15 \Rightarrow dy = 2 dx\) The converted integrals are: \(A = \int_{5}^{9} 1/2 dy + \int_{9}^{15} 2 dx\)

Step 3: 3. Compute the integrals

Now, compute the integrals: First integral: \(\int_{5}^{9} \frac{1}{2} dy = \frac{1}{2}(y \Big|_{5}^{9})=\frac{1}{2}(9-5) = 2\) Second integral: \(\int_{9}^{15} 2 dx = 2(x \Big|_{9}^{15})=2(15-9) = 12\)

Step 4: 4. Add the individual areas to find the final area

Finally, add the areas from both integrals together: \(A = 2 + 12 = 14\) Therefore, the area of the region bounded by \(y=\sqrt{x}, y=2x-15\) and \(y=0\) is 14 square units.

Key Concepts

Integration

Integration is an essential tool in calculus used to find quantities such as areas, volumes, and other accumulated values. It can be thought of as the reverse process of differentiation. Whereas differentiation involves finding the rate at which a function changes, integration is about summing up small amounts (or rectangles) to find a total amount, like area or volume.

• If you had a graph of a function, integration could help calculate the area under the curve.
• The integral sign \( \int \) is used to denote integration.
• The most common type of integration we deal with is called definite integration.

Expanding on definite integrals, when the lower and upper limits are specified, the result of integration provides a specific value representing the area under the curve between those limits. These are crucial when computing the area between curves, as in our example problem.

Area between curves

The area between two curves is one of the applications of definite integration. This is usually solved by integrating the top function minus the bottom function, over the interval created by their points of intersection.

• To effectively compute the area between two curves, you first determine the intersection points, as they set the boundaries.
• The formula to calculate this area involves finding the difference between the upper curve and the lower curve, integrated over the interval of intersection.

In the example given, the curves are \( y = \sqrt{x} \) and \( y = 2x - 15 \), alongside \( y = 0 \). You split the integration based on where the curves intersect. This is validated by determining which curve is above or below within different sections, bounded by their intersection points.

Definite integrals

Definite integrals are used to find the net area bounded by a curve and the x-axis over a certain interval. The value obtained is the difference of two related function values at specified limits of integration. Definite integrals take the form \( \int_{a}^{b} f(x) \, dx \).

• Here, \(a\) and \(b\) are the lower and upper limits of integration, marking the interval to be considered.
• The result is a number, representing the accumulated value of the function over this interval, often an area.

In the problem at hand, the definite integrals computed are based on transformed expressions, since the original curves were given in terms of \(y\). By converting them to variables of \(x\), their areas can be calculated accordingly for distinct segments, ensuring an accurate result.

Intersection points

Finding the intersection points of curves is a crucial initial step in problems involving areas between curves. These points help define the limits for integration.

• Intersection points occur where the curves meet—meaning the functions are equal for those specific values of \(x\) or \(y\).
• The process often involves equating the equations of the curves and solving for the variable.
• Once found, these points enable you to set up the correct intervals for calculating area.

For the given problem, solving \( y = \sqrt{x} \) and \( y = 2x - 15 \) yields two intersection points, \((5, -5)\) and \((9, 3)\). These are determined by equating the equations for \(y\), leading to a simplified one-variable quadratic equation, solved to find the values of \(y\), and subsequently, \(x\).