Problem 68 - 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 that 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 \(\frac{5}{2}\).

Step 1: Utilize the property of even functions

Since both \(f(x)\) and \(g(x)\) are even integrable functions, it means that \(\int_{-a}^a f(x) dx = 2\int_0^a f(x) dx\) and \(\int_{-a}^a g(x) dx = 2\int_0^a g(x) dx\).

Step 2: Area equation with respect to integration

We know that the area bounded by the graphs of \(f\) and \(g\) on \([-a,a]\) is 10. To translate that into integration, we know that \(\int_{-a}^a [f(x)-g(x)]dx= 10\).

Step 3: Use even function property to rewrite area equation

Now we can rewrite the integral equation from Step 2 utilizing the even function property from Step 1: \(2\int_0^a (f(x) - g(x)) dx = 10\).

Step 4: Solve for the integral \(\int_{0}^a [f(x)-g(x)] dx\)

From the equation we found in Step 3, we can solve for the integral: \(\int_{0}^a [f(x)-g(x)] dx = \frac{10}{2}=5\).

Step 5: Perform a change of variables with \(u = x^2\)

To solve the given integral \(\int_{0}^{\sqrt{a}} x[f(x^2) - g(x^2)] dx\), we will substitute \(u = x^2 \Rightarrow x =\sqrt{u}\). For the derivative, we have \(\frac{du}{dx} = 2x\), so \(dx = \frac{du}{2x}\). Also, the limits will change: when \(x=0\), \(u=0\) and when \(x=\sqrt{a}\), \(u=a\).

Step 6: Rewrite the integral with the change of variables

Using the change of variables, we can rewrite the integral as follows: \(\int_{0}^{a} \sqrt{u}[f(u) - g(u)] \frac{du}{2\sqrt{u}}\).

Step 7: Simplify the integral and find the value

We can simplify the rewritten integral by canceling out the \(\sqrt{u}\), and taking the constant factor 1/2 out: \(\frac{1}{2}\int_{0}^{a} [f(u) - g(u)] du\). Now, since we know the value of \(\int_{0}^a [f(x)-g(x)] dx=5\), just substitute: \(\frac{1}{2} \cdot 5 = \frac{5}{2}\). So, the value of \(\int_{0}^{\sqrt{a}} x[f(x^2) - g(x^2)] dx = \boxed{\frac{5}{2}}\).

Key Concepts

Even Functions

Even functions hold unique mathematical properties, making them particularly useful in integral calculus. A function \( f(x) \) is considered even if it satisfies the condition \( f(x) = f(-x) \) for all \( x \) in its domain. This symmetry about the y-axis is quite helpful when dealing with definite integrals over symmetric intervals.
For even functions, the integral from \( -a \) to \( a \) can be simplified. Instead of calculating the integral over both the negative and positive intervals, we can compute from \( 0 \) to \( a \) and multiply the result by 2. So, for any even function \( f(x) \), we have:

• \( \int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx \)

This property dramatically reduces the complexity of calculations, providing a convenient shortcut in problem-solving. It helps us simplify and solve the exercise by exploiting this symmetry.

Integration by Substitution

Integration by substitution is a technique similar to the reverse chain rule used in differentiation. It simplifies integrals by transforming them into a more easily solvable form.
To apply integration by substitution in an exercise, we follow these steps:

• Identify a substitution variable \( u \), usually chosen so that the integral becomes simpler.
• Replace the original variable and differential \( dx \) with \( du \) using the chosen substitution.
• Change the integral's limits if definite integration is performed.

In the provided exercise, the substitution \( u = x^2 \) was made, meaning \( x = \sqrt{u} \) and \( dx = \frac{du}{2x} \). This transformation alters the original integral into a simpler form and is crucial in moving further towards finding the solution. Integration by substitution is invaluable in simplifying complex integrals.

Definite Integrals

Definite integrals calculate the net area under a curve between two points \( a \) and \( b \). The result is determined across this interval and capturing the area between the function and the x-axis.
The solution to a definite integral of a function \( f(x) \) over \([a, b]\) is given by:

• \( \int_a^b f(x) \, dx = F(b) - F(a) \), where \( F(x) \) is an antiderivative of \( f(x) \)

This captures the accumulated change from \( a \) to \( b \). In the exercise, the definite integrals of even functions and substituted variables are used to solve for specific areas bound by the graphs, simplifying the computation.
Using the properties of even functions and substitution helps in calculating definite integrals more efficiently, leveraging symmetry and algebraic transformation.