Question

An integral equation Use the Fundamental Theorem of Calculus, Part \(1,\) to find the function \(f\) that satisfies the equation $$ \int_{0}^{x} f(t) d t=2 \cos x+3 x-2 $$ Verify the result by substitution into the equation.

Short answer

Answer: The function f(t) that satisfies the given integral equation is f(t) = -2 sin x + 3.

Step-by-step solution

Write the integral equation

We have the following integral equation: $$ \int_{0}^{x} f(t) dt = 2 \cos x + 3x - 2 $$

Use the Fundamental Theorem of Calculus, Part 1

According to the Fundamental Theorem of Calculus, Part 1, if F(x) is an antiderivative of f(x) on the interval [0, x], then the following equation holds: $$ \int_{0}^{x} f(t) dt = F(x) - F(0) $$ Now, differentiate both sides of the integral equation with respect to x to obtain the function f(x): $$ \frac{d}{dx} \int_{0}^{x} f(t) dt = \frac{d}{dx} (2 \cos x + 3x - 2) $$ By the Fundamental Theorem of Calculus, the left side is just f(x): $$ f(x) = -2 \sin x + 3 $$

Verify the result through substitution

To verify the result, we will find the antiderivative of our obtained f(x) and check whether it satisfies the given integral equation. The antiderivative of f(x) = -2 sin x + 3 is: $$ F(x) = -2 \int \sin x \, dx + 3 \int 1 \, dx = 2 \cos x + 3x + C $$ Now, we substitute the value of F(x) in the equation we obtained in Step 2: $$ \int_{0}^{x} f(t) dt = F(x) - F(0) = (2 \cos x + 3x + C) - (2 \cos 0 + 3 \cdot 0 + C) = 2 (\cos x - 1) + 3x $$ Therefore, we find that the integral of our derived f(t) satisfies the given equation: $$ \int_{0}^{x} f(t) dt = 2 \cos x + 3x - 2 $$ So the function f(t) = -2 sin x + 3 is indeed the correct solution.

Key concepts

Integral Equation

An integral equation is a relationship that involves an unknown function inside an integral. In our given problem, the equation \(\int_{0}^{x} f(t) dt = 2 \cos x + 3x - 2\) is such an equation. The task is to determine the function \(f(t)\) whose definite integral from 0 to \(x\) equals a specified expression.
Integral equations can often be solved by finding derivatives. This approach ties into a fundamental concept in calculus—the Fundamental Theorem of Calculus, which connects derivatives and integrals in a neat loop. We use differentiation to discover the function \(f(x)\) that will fulfill the given condition.

Antiderivative

An antiderivative of a function \(f(x)\) is another function \(F(x)\) such that the derivative of \(F(x)\) equals \(f(x)\). In mathematical notation, if \(F'(x) = f(x)\), then \(F(x)\) is an antiderivative of \(f(x)\).
When solving an integral equation, finding the antiderivative is crucial. As per the problem, once we determine \(f(x) = -2 \sin x + 3\) by differentiating the given expression, we check for the antiderivative. By integrating \(-2 \sin x + 3\), we get \(F(x) = 2 \cos x + 3x + C\). Here, \(C\) represents the constant of integration, an arbitrary constant added since derivatives of constants are zero. Confirming the correctness involves substituting back into the original equation to check if both sides of the equation coincide.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function, or the rate at which it changes. It serves as one of the main tools to solve integral equations. According to the Fundamental Theorem of Calculus, the derivative of the integral of a function \(f(t)\) from a constant value to \(x\) is simply \(f(x)\).
In the exercise we have \(\int_{0}^{x} f(t) dt = 2 \cos x + 3x - 2\). By differentiating both sides, we turn the process of integration into solving for the function \(f(x)\). We thus find \(f(x) = -2 \sin x + 3\) by differentiating \(2 \cos x + 3x - 2\). This step both simplifies the original integral equation and finds the required function that satisfies it.