Question

The area under the curve \(y=\sqrt{x}\) bounded from below by the \(x\) -axis and on the right by \(x=4\) is \(\frac{16}{3}\) square units. Using the ideas presented in this section, what do you think is the area under the curve of \(y=\sqrt{-x}\) bounded from below by the \(x\) -axis and on the left by \(x=-4 ?\) Justify your answer.

Short answer

The area is \( \frac{16}{3} \) square units due to symmetry.

Step-by-step solution

Understanding the given problem

The problem provides the area under the curve of the function \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 4 \) as \( \frac{16}{3} \) square units. The task is to find the area under the curve of the function \( y = \sqrt{-x} \) from \( x = -4 \) to \( x = 0 \).

Analyzing the symmetry of the functions

Note that \( y = \sqrt{x} \) and \( y = \sqrt{-x} \) are symmetric about the y-axis. Specifically, in this exercise, the function \( y = \sqrt{-x} \) is just the reflection of the function \( y = \sqrt{x} \) across the y-axis.

Setting up the integral for \( y = \sqrt{-x} \)

To find the area under the curve \( y = \sqrt{-x} \) from \( x = -4 \) to \( x = 0 \), set up the integral of \( y = \sqrt{-x} \) within the given bounds. The integral will be: \[ \int_{-4}^{0} \sqrt{-x} \, dx. \]

Substituting and solving the integral

Make the substitution \( u = -x \). Then, \( du = -dx \), and the integral becomes: \[ -\int_{4}^{0} \sqrt{u} \, du. \] Changing the integration limits back to make it easier: \[ \int_{0}^{4} \sqrt{u} \, du. \]

Evaluating the integral

Evaluate the integral: \[ \int_{0}^{4} u^{1/2} \, du = \left[ \frac{2u^{3/2}}{3} \right]_0^4 = \frac{2(4^{3/2})}{3} - 0 = \frac{2(8)}{3} = \frac{16}{3}. \] So the area under the curve \( y = \sqrt{-x} \) from \( x = -4 \) to \( x = 0 \) is also \( \frac{16}{3} \) square units.

Confirming the symmetry

Since \( y = \sqrt{x} \) from \( x = 0 \) to \( x = 4 \) has the same area as \( y = \sqrt{-x} \) from \( x = -4 \) to \( x = 0 \), the given area relationship is confirmed by symmetry.

Key concepts

Integral Calculation

To find the area under a curve, we use the concept of integration. Integrals help us sum up an infinite number of infinitesimally small areas under a curve. In this exercise, we are given two functions, \(y = \sqrt{x}\) and \(y = \sqrt{-x}\), and we need to find the area under these curves over specified intervals. For the function \(y = \sqrt{x}\), the area from \(x = 0\) to \(x = 4\) is already given as \(\frac{16}{3}\) square units. Now, we need to find the area for the function \(y = \sqrt{-x}\) from \(x = -4\) to \(x = 0\). By setting up the integral \(\int_{-4}^{0} \sqrt{-x} \ dx\) and solving it with appropriate substitution, we determine this area. This fundamental integral calculation technique lets us generalize areas for various functions.

Function Reflection

The function \(y = \sqrt{x}\) and its reflection across the y-axis, \(y = \sqrt{-x}\), form symmetric pairs. Reflecting a function means flipping it over a specific axis – here, the y-axis. This symmetry indicates that the area calculation on one side of the axis will mirror the area on the other side. In our problem, because \(y = \sqrt{x}\) has a specific area in the first quadrant, \(y = \sqrt{-x}\) will have the same area in the second quadrant. This is because the operations performed on \(x\) and \(-x\) alter the direction but not the magnitude of the integral evaluation. This symmetry greatly simplifies the problem, saving calculation effort and ensuring easy verification of results.

Definite Integral

A definite integral calculates the net area under a curve within given boundaries. In this tutorial, we set up the definite integral for \(y = \sqrt{-x}\) over the interval \(x = -4\) to \(x = 0\): \( \int_{-4}^{0} \sqrt{-x} \ dx \). By applying the substitution \( u = -x \), which transforms negative limits into positive, this definite integral converts to \( \int_{0}^{4} \sqrt{u} \ du \). Evaluating this standard integral, we find the area under the curve using fundamental integration rules. The process involves solving the integral, applying the Fundamental Theorem of Calculus, and evaluating the result at the given boundaries, giving us an explicit value for the area.

Symmetric Functions

Symmetric functions are a powerful concept in calculus. A function is symmetric around an axis if flipping it across this axis leaves the function graph unchanged. In this lesson, we see that \(y = \sqrt{x}\) and \(y = \sqrt{-x}\) are symmetric about the y-axis. This means their areas under the curve within mirrored intervals are identical. Understanding these symmetries simplifies computations, as we can infer the solutions for one part of the curve from the other. With areas under symmetric functions frequently equal, this property often saves time and computational effort when solving integral problems. Thus, recognizing symmetry in functions is invaluable for efficient problem-solving in calculus.