Problem 68 - Calculus Single Variable

Question

Integrate by parts. This will result in an integrand of the form \(P(x) / Q(x)\) where \(P(x)\) and \(Q(x)\) are polynomials with the degree of \(P(x)\) greater than or equal to the degree of \(Q(x)\). Such an integrand is handled by performing polynomial division to put \(P(x) / Q(x)\) into the form \(r(x)+s(x) / Q(x)\) where \(r(x)\) and \(s(x)\) are polynomials with the degree of \(s(x)\) less than the degree of \(Q(x)\) $$ \int_{0}^{1} x \ln (1+x) d x $$

Step-by-Step Solution

Answer

The integral \( \int_{0}^{1} x \ln(1+x) \, dx \) evaluated results in \( \frac{1}{4} - \frac{1}{9} \).

Step 1: Choose functions for integration by parts

Integration by parts relies on the formula \( \int u \, dv = uv - \int v \, du \). We choose \( u = \ln(1+x) \) because its derivative is simpler, and \( dv = x \, dx \), making it straightforward to integrate.

Step 2: Differentiate and integrate terms

Compute \( du \) and \( v \): - Differentiate \( u = \ln(1+x) \) to get \( du = \frac{1}{1+x} \, dx \).- Integrate \( dv = x \, dx \) to get \( v = \frac{x^2}{2} \).

Step 3: Apply integration by parts formula

Substitute into the integration by parts formula:\[ \int x \ln(1+x) \, dx = \left[ \ln(1+x) \cdot \frac{x^2}{2} \right] - \int \frac{x^2}{2} \cdot \frac{1}{1+x} \, dx \].Evaluate jointly within the limits from 0 to 1.

Step 4: Simplify the integral

Simplify the expression:- The first term evaluates as: \( \left. \frac{x^2}{2} \ln(1+x) \right|_0^1 \).- The integral simplifies to: \[ \int \frac{x^2}{2(1+x)} \, dx \].

Step 5: Perform polynomial division

Perform polynomial division on \( \frac{x^2}{2(1+x)} \). Divide \( x^2 \) by \( 1+x \) to yield \( x - x^2 \) and a remainder \( x^2 \). Integrate \( x - x^2 + \frac{-x^2}{1+x} \).

Step 6: Final integration and evaluation

Integrate each component obtained:- \( \int x \ dx = \frac{x^2}{2}, \int x^2 \ dx = \frac{x^3}{3}, \int \frac{-x^2}{1+x} \ dx \) leads to simpler forms.Evaluate from 0 to 1, combine results and calculate differences.

Step 7: Evaluate the definite integral

Apply limits to the antiderivative found after integration: \[ \left[ \frac{x^2}{2} \ln(1+x) - \left( \frac{x^2}{2} - \frac{x^3}{3} \right) \right]_0^1 \]. Solve for the final result.

Step 8: Confirm solution

Calculate the terms separately for the range from 0 to 1: - First term at 1 minus term at 0. Simplify the result to conclude.

Key Concepts

Polynomial Division

Polynomial division is a technique used to simplify expressions where one polynomial is divided by another. This method is particularly useful when dealing with integration problems that result in fractional expressions where the degree of the numerator is greater than or equal to the degree of the denominator.

When you perform polynomial division, you start by dividing the leading term of the numerator by the leading term of the denominator. This gives you the first term of the quotient. You then multiply this term by the entire divisor, and subtract the result from the original numerator.

Repeat this process with the new numerator obtained after subtraction, until the degree of the new numerator is less than the degree of the original denominator. The quotient obtained is the simplified integer polynomial, and any leftover expression is the remainder.

This remainder is then expressed over the original denominator to complete the conversion of the fraction into a simpler form. This process is vital in preparing an expression for integration by separating the polynomial part from the fractional part.

Definite Integrals

Definite integrals are a fundamental concept in calculus that involve integrating a function over a specific interval. The result of a definite integral is a numerical value that represents the area under the curve of the function within the given limits.

The primary difference between definite and indefinite integrals is that definite integrals have specified limits of integration. These limits are denoted at the top and bottom of the integral sign and are typically in the form \( \int_{a}^{b} f(x) \, dx \). This notation indicates that you need to integrate the function \( f(x) \) from \( a \) to \( b \).

When solving a definite integral, you first find the antiderivative of the function. Then, you apply the Fundamental Theorem of Calculus, which involves evaluating the antiderivative at the upper limit of integration and subtracting the antiderivative evaluated at the lower limit.

This gives a precise way to calculate the net area between the curve and the x-axis within the given interval. It is a powerful tool in analyzing the total accumulation of quantities, changes, or even physical areas in various fields.

Antiderivatives

An antiderivative is essentially the reverse process of differentiation. It involves finding a function whose derivative yields the original function. In other words, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F'(x) = f(x) \).

Antiderivatives are crucial in calculus because they form the basis for calculating integrals. When we integrate a function, we are actually seeking its antiderivative. The general form of an antiderivative is represented by the integral symbol \( \int f(x) \, dx \), and often includes a constant of integration \( C \), since differentiation of constants yields zero.

In the specific case of definite integrals, we find that the constant of integration cancels out, as we are evaluating the definite boundaries and looking for the net difference.

Overall, understanding antiderivatives is essential for solving both definite and indefinite integrals, as it allows you to bridge the gap between the problem at hand and finding a solution that addresses the total area or accumulation as desired.