Problem 72 - Calculus Early Transcendentals

Question

Assume \(f\) and \(g\) are even, integrable functions on \([-a, a],\) where \(a>1 .\) Suppose \(f(x)>g(x)>0\) on \([-a, a]\) and the area bounded by the graphs of \(f\) and \(g\) on \([-a, a]\) is \(10 .\) What is the value of \(\int_{0}^{\sqrt{a}} x\left(f\left(x^{2}\right)-g\left(x^{2}\right)\right) d x ?\)

Step-by-Step Solution

Answer

Answer: The value of the integral is \(2.5\).

Step 1: Find the integral for the area between the functions

Since we are given the area between the two functions is \(10\), we can represent this area using the integral \(\int_{-a}^{a} \left(f(x) - g(x)\right) \mathrm{d}x = 10.\) As \(f\) and \(g\) are even functions, we can apply the property for even functions to find the area from \(0\) to \(a\): $$2\int_{0}^{a} \left(f(x) - g(x)\right) \mathrm{d}x = 10.$$

Step 2: Divide by 2

As given above, we now have the equation \(2\int_{0}^{a} \left(f(x) - g(x)\right) \mathrm{d}x = 10\). Divide by \(2\) to get the integral from \(0\) to \(a\): $$\int_{0}^{a} \left(f(x) - g(x)\right) \mathrm{d}x = 5.$$

Step 3: Rewrite the integral with substitution

In order to rewrite the integral, \(\int_{0}^{\sqrt{a}} x\left(f\left(x^{2}\right)-g\left(x^{2}\right)\right) \mathrm{d}x\), we will apply the substitution method. Let's define the variable \(u = x^2\). In this case, \(\frac{d}{dx}u = 2x\Rightarrow \mathrm{d}u = 2x\mathrm{d}x\). Substitute this into the integral (also change the bounds by substituting the original bounds in the new variable; When \(x = \sqrt{a}\), \(u = a\) and when \(x = 0\), \(u = 0\) ): $$\int_{0}^{a} \frac{1}{2}\left(f(u)-g(u)\right) \mathrm{d}u.$$

Step 4: Compare integrals and find the answer

Now, we can see a relationship between the result of Step 2 and the integral from Step 3. Our equation from step 2 is: $$\int_{0}^{a} \left(f(x) - g(x)\right) \mathrm{d}x = 5,$$ and our integral from step 3 is: $$\int_{0}^{a} \frac{1}{2}\left(f(u)-g(u)\right) \mathrm{d}u.$$ By comparing these two expressions, we can conclude that the integral from step 3 is half the integral from step 2: $$\int_{0}^{a} \frac{1}{2}\left(f(u)-g(u)\right) \mathrm{d}u = \frac{1}{2}\int_{0}^{a} \left(f(x) - g(x)\right) \mathrm{d}x.$$ Now, since we know from step 2 that \(\int_{0}^{a} \left(f(x) - g(x)\right) \mathrm{d}x = 5,\) we can find the value of the integral in step 3: $$\int_{0}^{a} \frac{1}{2}\left(f(u)-g(u)\right) \mathrm{d}u = \frac{1}{2} \cdot 5 = 2.5.$$ Thus, the value of the integral \(\int_{0}^{\sqrt{a}} x\left(f\left(x^{2}\right)-g\left(x^{2}\right)\right) \mathrm{d}x\) is \(\boxed{2.5}\).

Key Concepts

Even Functions

Even functions have special properties that simplify calculations in definite integrals, especially over symmetric intervals like \([-a, a]\). An even function, by definition, satisfies the condition \(f(-x) = f(x)\) for all x within its domain. This symmetry results in certain simplifications:

• When integrating an even function over a symmetric interval \([-a, a]\), it can be reduced to twice the integral over the interval \[0, a\].
• This simplification is expressed mathematically as \[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \].

In the original exercise, both \(f\) and \(g\) are even functions, allowing us to apply this property. This property aids in calculating the area between the curves by reducing computations to one part of the symmetric interval.

Substitution Method

The substitution method is a clever technique used in calculus to simplify integrals, often transforming them into a more solvable form. The idea is to choose a new variable that simplifies the integrand. Here’s how it typically works:

• Identify a substitution that simplifies the integration, like setting \(u = x^2\).
• Then, calculate the differential, \(\text{d}u = 2x \, \text{d}x\), reflecting how infinitesimal changes in \(x\) relate to changes in \(u\).
• Adjust the limits of integration according to your substitution. For instance, when \(x = 0, \, u = 0\) and when \(x = \sqrt{a}, \, u = a\).

The real benefit of substitution is the ability to change a complex integral into a more manageable one. In the problem, by using the substitution \(u = x^2\), we transformed the given integral into a simpler form, making it easier to compare and solve.

Area between Curves

Understanding the area between curves is essential in calculus. This concept involves identifying the region between two functions over a specific interval:

• Given two functions \(f(x)\) and \(g(x)\), the area between them is determined by integrating the difference \(f(x) - g(x)\) from \(a\) to \(b\).
• Visually, this represents the vertical space between their graphs across the x-axis span.
• In many instances, the functions involved are even, simplifying the integral as shown in the previous section.

In the exercise, with \(f(x) > g(x) > 0\), we explored the integral \((f(x) - g(x))\) over \([-a, a]\) known to be 10. These steps are vital for efficiently calculating the exact area trapped between the two functions over a symmetric region and ultimately solving the problem's integral expression by breaking it into more manageable parts.