Problem 48 - Calculus Single Variable

Question

Suppose that \(f\) is continuous on \((-a, a)\) and that \(f\) is an even function: \(f(-t)=f(t) .\) Differentiate \(F(x)=\int_{0}^{x} f(t) d t-\) \(\int_{-x}^{0} f(t) d t\) with respect to \(x,\) and use your result to show that \(\int_{-x}^{0} f(t) d t=\int_{0}^{x} f(t) d t\).

Step-by-Step Solution

Answer

The equality \( \int_{0}^{x} f(t) \, dt = \int_{-x}^{0} f(t) \, dt \) is shown by differentiating and using the even nature of \( f \), resulting in a constant \( F(x) \).

Step 1: Express Given Integrals

The given function is: \[ F(x) = \int_{0}^{x} f(t) \, dt - \int_{-x}^{0} f(t) \, dt \]

Step 2: Apply Leibniz Rule for Differentiation

Differentiate the integrals with respect to \(x\). Using Leibniz's rule:\[ \frac{d}{dx} \left( \int_{0}^{x} f(t) \, dt \right) = f(x) \] Similarly,\[ \frac{d}{dx} \left( -\int_{-x}^{0} f(t) \, dt \right) = -f(-x) \cdot (-1) = f(-x) \]

Step 3: Differentiate F(x)

Now differentiate \( F(x) \):\[ F'(x) = \frac{d}{dx} \left( \int_{0}^{x} f(t) \, dt \right) - \frac{d}{dx} \left( \int_{-x}^{0} f(t) \, dt \right) \]Substitute the results from Leibniz's rule:\[ F'(x) = f(x) - f(-x) \]

Step 4: Simplify Expression Using Even Function Property

Since \( f \) is an even function, \( f(x) = f(-x) \). Therefore, substitute this into the derivative of \( F(x) \):\[ F'(x) = f(x) - f(x) = 0 \]

Step 5: Analyze Constant Derivative Result

Since \( F'(x) = 0 \), it implies that \( F(x) \) is a constant function.

Step 6: Conclude Integral Equality Using Constant F(x)

Since \( F(x) = 0 \) for all \( x \), the integrals are equal:\[ \int_{0}^{x} f(t) \, dt = \int_{-x}^{0} f(t) \, dt \]

Key Concepts

Continuous Function

A continuous function is a function that doesn't have any interruptions, jumps, or breaks in its domain. In mathematical terms, a function \( f \) is continuous on an interval if, for every value \( c \) in that interval, the limit of \( f(x) \) as \( x \) approaches \( c \) is equal to \( f(c) \). This means that as you get closer to \( c \), the values of the function get closer to the output value at \( c \).Continuous functions have no sudden changes in value, which is important when performing operations like differentiation and integration. In the given exercise, we have the continuous function \( f \) on the interval \((-a, a)\), ensuring smooth behavior across this interval. This consistency becomes especially crucial when applying integral calculus, as it allows us to work with predictable and manageable functions.

Even Function

An even function is a function that is symmetric with respect to the y-axis, meaning its graph looks the same if flipped over the y-axis. Mathematically, a function \( f \) is defined as even if for every real number \( t \), \( f(-t) = f(t) \). This symmetry simplifies many calculations in calculus, especially in the context of integrals and derivatives.In the exercise, the function \( f \) is given as even, which allows us to simplify expressions involving \( f \). When finding the derivative of the function \( F(x) \), we use the property of even functions to conclude that \( F'(x) = f(x) - f(x) \), which equals zero. This property of even functions greatly helps in proving integral equality, as it aids in reducing complexity.

Leibniz Rule

Leibniz's rule, named after mathematician Gottfried Wilhelm Leibniz, provides a method for differentiating under the integral sign. This rule is particularly useful when a function of some variables is defined by an integral. Its formula simplifies the process of differentiating when the limits of the integral are functions of the variable of differentiation.In the solution to our exercise, Leibniz's rule is applied to differentiate each of the integrals in \( F(x) = \int_{0}^{x} f(t) \, dt - \int_{-x}^{0} f(t) \, dt \). Applying the rule, the differentiation of \( \int_{0}^{x} f(t) \, dt \) with respect to \( x \) gives \( f(x) \), and differentiation of \( -\int_{-x}^{0} f(t) \, dt \) results in \( f(-x) \). This application of Leibniz's rule helps to simplify our solution and arrive at the derivative \( F'(x) = 0 \).

Definite Integrals

A definite integral calculates the accumulation of quantities, which can be thought of as the area under the curve of a function \( f(t) \) over a given interval \([a, b]\). When computing a definite integral, both the start and the end points are specified, making the result a constant value rather than a function.In this problem, we deal with the definite integrals \( \int_{0}^{x} f(t) \, dt \) and \( \int_{-x}^{0} f(t) \, dt \). By discovering that \( F(x) \) is a constant function due to the derivative \( F'(x) = 0 \), it shows that the two definite integrals provide the same accumulated value of \( f(t) \) over their respective intervals. This equality highlights a beneficial property of symmetric (even) functions and how their integrals mirror over the y-axis.