Problem 11 - Calculus Single Variable

Question

Suppose that \(\int_{1}^{4} f(x) / 3 d x=2 .\) Evaluate \(\int_{4}^{1} 3 f(x) d x\).

Step-by-Step Solution

Answer

\(\int_{4}^{1} 3f(x) \, dx = -6\."

Step 1: Understand the Problem

We are given an integral \( \int_{1}^{4} \frac{f(x)}{3} \, dx = 2 \) and need to find \( \int_{4}^{1} 3f(x) \, dx \). Notice the limits of integration are reversed in the second integral.

Step 2: Use the Property of Integrals

Recall that reversing the limits of an integral changes its sign: \( \int_{a}^{b} g(x) \, dx = -\int_{b}^{a} g(x) \, dx \). Apply this property to the given integral: \( \int_{1}^{4} \frac{f(x)}{3} \, dx = -\int_{4}^{1} \frac{f(x)}{3} \, dx = 2 \). Thus, \( \int_{4}^{1} \frac{f(x)}{3} \, dx = -2 \).

Step 3: Convert to Desired Integral

Multiply both sides of the equation \( \int_{4}^{1} \frac{f(x)}{3} \, dx = -2 \) by 3 to find \( \int_{4}^{1} 3f(x) \, dx \). This gives \( 3 \times \int_{4}^{1} \frac{f(x)}{3} \, dx = 3(-2) \).

Step 4: Solve for the Desired Integral

Compute the right side of \( 3 \times (-2) = -6 \). Thus, \( \int_{4}^{1} 3f(x) \, dx = -6 \).

Key Concepts

Integration by parts

The technique of Integration by Parts is a vital method in Integral Calculus to simplify complex integrals. It's particularly useful when dealing with the product of two functions. The fundamental principle is derived from the product rule of differentiation.
The formula is presented as follows:

• If you have an integral of the form \( \int u \, dv \), it can be transformed to \( uv - \int v \, du \).

Here’s a breakdown:

• \( u \): Choose the first function to differentiate. It could be a polynomial, logarithmic, or inverse trigonometric function.
• \( dv \): Choose the second function to integrate. This is the remaining part of the integral.

Apply this method when integrating products of functions, like a polynomial times a sine or cosine function. By carefully selecting which part of the integrand becomes \( u \) and which becomes \( dv \), this approach effectively reduces complex integrals into simpler forms.
Remember, the power of Integration by Parts lies in its ability to tackle otherwise difficult integrations by breaking them into more manageable calculations.

Definite integrals

Definite integrals are a core concept in calculus, providing the area under a curve within specific bounds. The notation \( \int_{a}^{b} f(x) \, dx \) expresses a definite integral from \( a \) to \( b \).
Key characteristics include:

• Limits of Integration: The values \( a \) and \( b \) set the interval over which the area under the function \( f(x) \) is calculated.
• Sign Change on Limits: If you swap \( a \) and \( b \), the integral's value changes sign. This property is critical and was used in the solution to reverse the original integral's sign and solve the exercise.

Definite integrals are not just about finding areas; they're instrumental in solving physics and engineering problems, where quantities are measured over intervals.
Applications range from calculating displacement when given a velocity function, to finding the total accumulated change where variables evolve across defined intervals.

Properties of integrals

Integrals come with a set of properties that simplify computations and problem-solving in integral calculus. Understanding and utilizing these properties can make tackling complex integrals much more straightforward.
A few key properties include:

• Linearity: If \( c \) is a constant, then \( \int c f(x) \, dx = c \int f(x) \, dx \). This property was crucial in the solution provided, allowing for multiplication of both sides of an integral by 3.
• Reversal of Limits: Swapping the bounds of integration in an integral, such as turning \( \int_{a}^{b} f(x) \, dx \) into \( \int_{b}^{a} f(x) \, dx \), inverts the sign of the integral. This property was used to transform \( \int_{1}^{4} \frac{f(x)}{3} \, dx \) in the exercise.

These properties streamline the integration process and provide shortcuts that allow complicated integral expressions to be broken down into simpler, manageable forms. Mastering them is essential for efficiently solving a wide range of calculus problems.