Problem 56 - Calculus Single Variable

Question

The integral \(\int_{a}^{b}\left(f_{1}(x)-f_{2}(x)\right) d x\) represents the area of a region in the \(x y\) -plane that is bounded by the graphs of \(f_{1}\) and \(f_{2}\). Express the area of the region as an integral of the form \(\int_{c}^{d}\left(g_{1}(y)-g_{2}(y)\right) d y .\) For example, the integral \(\int_{0}^{1}\left(x-x^{2}\right) d x\) represents the area of the shaded region in Figure \(11 .\) This area can also be represented as \(\int_{0}^{1}(\sqrt{y}-y) d y.\) $$ \int_{-1}^{1}(e-\exp (|x|)) d x $$

Step-by-Step Solution

Answer

The equivalent integral in terms of y is \( \int_{1}^{e} 2 \ln(y) \, dy \).

Step 1: Understand the Integral in Terms of Area

The given integral is \( \int_{-1}^{1}(e - \exp(|x|)) \, dx \). This represents the area between the curves \( y = e \) and \( y = \exp(|x|) \) from \( x = -1 \) to \( x = 1 \). Since \( e \) is a constant function and \( \exp(|x|) \) is symmetric around the y-axis, the region between them is bounded above by \( y = e \) and below by \( y = \exp(|x|) \).

Step 2: Convert the Functions to Express x in Terms of y

To represent the area using integration with respect to \( y \), we need to determine \( x \) as a function of \( y \). Start with the equation \( y = \exp(|x|) \). Solving gives \( |x| = \ln(y) \), meaning \( x = \pm \ln(y) \), since \( x \) could be negative or positive.

Step 3: Identify the New Integration Bounds

The range of \( y \) values is from the minimum of \( \exp(|x|) \) at \( x = 0 \) to \( e \) (since the functions intersect when \( \exp(|x|) = e \)). Therefore, as \( |x| \) grows from \( 0 \) to \( 1 \), \( y \) goes from \( e^0 = 1 \) to \( e^1 = e \). Hence, the bounds for \( y \) are \( 1 \) to \( e \).

Step 4: Define the Functions in Terms of y

For a fixed value of \( y \), the curve \( \exp(|x|) = y \) corresponds to \( x = -\ln(y) \) on the left of the y-axis and \( x = \ln(y) \) on the right of the y-axis. Therefore, the horizontal distance between these two curves is \( \ln(y) - (-\ln(y)) = 2\ln(y) \).

Step 5: Formulate the Integral in Terms of y

With the new functions and bounds established, express the area as \( \int_{1}^{e} 2 \ln(y) \, dy \). This integral calculates the area between the vertical lines representing the x-values as functions of y and the constant y-value between 1 and e.

Key Concepts

Area under curves

The concept of the area under curves is crucial to understanding definite integrals. When you calculate the integral of a function over a certain interval, you are effectively finding the area between the curve of the function and the x-axis, over that interval. In more complex scenarios, like with the integral \( \int_{-1}^{1}(e - \exp(|x|)) \, dx \), you find the area between two curves—one above the other. Here, the area between \( y = e \) and \( y = \exp(|x|) \) is calculated.

• The curve \( y = e \) represents a horizontal line, because \( e \) is a constant.
• The curve \( y = \exp(|x|) \) is a symmetric curve that increases as \( x \) moves away from zero.

Calculating these areas manually by division into smaller shapes like rectangles is cumbersome, but integration provides a precise approach to obtaining these areas.

Change of variables

When shifting from integrating with respect to \( x \) to \( y \), we engage in a process called change of variables. This technique transforms the integral into one that may be easier to evaluate or interpret. By expressing variables in terms of another variable, you change the integration perspective.In the exercise, we start with \( y = \exp(|x|) \). Solving for \( x \) in terms of \( y \) gives us \( |x| = \ln(y) \) or \( x = \pm \ln(y) \).

• This allows us to convert the integral from a function of \( x \) to a function of \( y \).
• This transformation leads us to a new integral \( \int_{1}^{e} 2 \ln(y) \, dy \).

The change of variables often simplifies complex integrals by reducing them to easier forms or offering insights into symmetry or bounds.

Integration bounds

The integration bounds are the limits between which the function is evaluated and play a crucial role in determining the area. In \( \int_{-1}^{1}(e - \exp(|x|)) \, dx \), the bounds \(-1\) to \(1\) define the region over which we are calculating the area.When we convert our integral to respect \( y \), we need new bounds appropriate for this transformation. The integration bounds for \( y \) range from \(1\) to \(e\). This is because:

• The smallest value of \( y \) happens when \( \exp(|x|) = 1 \), here \( |x| = 0 \).
• The largest value of \( y \) is \( e \), occurring when \( |x| = 1 \).

These bounds encompass the entire region of interest, ensuring all relevant areas are accounted for in the new integral form.

Symmetry in integrals

Symmetry can simplify the evaluation of integrals by highlighting patterns and reducing computation. In the context of the integral \( \int_{-1}^{1}(e - \exp(|x|)) \, dx \), the function \( \exp(|x|) \) is symmetric about the y-axis. This symmetry means that the behavior of the function is mirrored on both sides of the y-axis.

• The integral is easier to evaluate because the area calculation may only need to be performed once (from 0 to 1) and then doubled, thanks to symmetry.
• Recognizing symmetry allows strategizing calculations either by reducing the interval or by performing clever transformations.

Thus, symmetry not only cuts down on calculations, but also provides insights into the nature of the function and the shape of the curve it describes.