Problem 37 - Calculus: Graphical, Numerical, Algebraic

Question

True or False If \(f^{\prime}(x)=g(x),\) then \(\int x^{2} g(x) d x=\) \(x^{2} f(x)-2 \int x f(x) d x .\) Justify your answer.

Step-by-Step Solution

Answer

True

Step 1: Identify Function u and dv

We can set \(u=x^2\) and \(v'=g(x)\) respectively. Then calculate \(du = 2x dx\) and \(v = ∫g(x) dx = f(x)\).

Step 2: Apply the formula of Integration by Parts

The formula \(\int u dv = u v - \int v du\) is applied here. Substituting \(u\), \(du\), \(v\) and \(dv\), we get \(\int x^2 g(x)dx = x^2 f(x) - \int f(x) * 2x dx\).

Step 3: Confirm the Equivalent Expression

We observe that the expression on the right hand side of the equation is the same as the one provided in the question. Hence, the statement is confirmed to be true.

Key Concepts

Integration by Parts

Integration by parts is a powerful technique used in calculus to integrate products of functions. The formula for integration by parts is given by:

• \( \int u \, dv = uv - \int v \, du \)

The goal is to transform a difficult integral into a simpler one. This method involves identifying parts of the function as \( u \) and \( dv \), then differentiating \( u \) to find \( du \) and integrating \( dv \) to find \( v \).

In our exercise, we've applied integration by parts by letting \( u = x^2 \) and \( dv = g(x) \, dx \). Consequently, the differential becomes \( du = 2x \, dx \) and \( v = \int g(x) \, dx = f(x) \). By substituting these into the formula, we get the original equation: \( \int x^2 g(x) \ dx = x^2 f(x) - 2 \int x f(x) \ dx \).

This confirms the expression is indeed an application of integration by parts.

Derivative

A derivative measures how a function changes as its input changes. If \( f(x) \) is a function, then \( f'(x) \) represents the derivative of \( f(x) \), also referred to as the rate of change or the slope of the function at any given point.

In the stated problem, it’s given that \( f'(x) = g(x) \). This means that \( g(x) \) is essentially the derivative of \( f(x) \). Thus, when you compute \( \int g(x) \, dx \), you retrieve the original function \( f(x) \), up to an integration constant.

• \( f'(x) = g(x) \)
• Therefore, \( \int g(x) \, dx = f(x) \)

Understanding derivatives is essential for many calculus operations, including integration, as they are inverse processes.

Definite Integral

The definite integral of a function represents the accumulation of quantities, such as areas under a curve. While our discussed problem involves an indefinite integral involving functions \( u \) and \( v \), definite integrals are crucial in many real-world applications where limits are specified.

It is expressed as \( \int_a^b f(x) \, dx \), representing the net area between the curve \( f(x) \) and the x-axis from \( x = a \) to \( x = b \).

In the exercise, we used the indefinite integral through the formula of integration by parts, but a good grasp of definite integrals complements your understanding of their estimates over specified intervals.

Functions

Functions are mathematical objects that establish a relationship between inputs and outputs. In calculus, understanding functions and their properties is key, as they form the basis of derivatives and integrals.

In the problem, two functions are involved: \( f(x) \) and \( g(x) \). They interact through their roles in differentiation and integration. The relationship \( f'(x) = g(x) \) hints that \( g(x) \) is derived from the function \( f(x) \), showing how these two functions are interconnected.

• \( f(x) \) and \( f'(x) \) explain the function's behavior and rate of change.
• \( g(x) \) as \( f'(x) \) indicates it's derived from \( f(x) \).

Grasping these relations is essential, as functions are foundational blocks in calculus, directing us on how to handle complex mathematical expressions like in the given problem.