Problem 23 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int s(2 s+1)^{4} d s\)

Step-by-Step Solution

Answer

\( \frac{s(2s+1)^5}{10} - \frac{(2s+1)^6}{120} + C \) is the integral solution.

Step 1: Identify Parts for Integration by Parts

Recall that the formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). We need to select what will be \( u \) and \( dv \). Let \( u = s \) and \( dv = (2s+1)^4 \, ds \).

Step 2: Differentiate and Integrate the Parts

Differentiate \( u = s \) to get \( du = ds \). Integrate \( dv = (2s+1)^4 \, ds \) by setting \( w = 2s+1 \) so \( dw = 2 \, ds \). Thus, \( ds = \frac{1}{2} dw \). Substitute to get \( \int w^4 \cdot \frac{1}{2} dw = \frac{1}{2} \int w^4 \cdot dw \).

Step 3: Integrate \( v \)

Compute \( \int w^4 dw \) which evaluates to \( \frac{w^5}{5} + C \). Substitute back \( w = 2s+1 \) to find \( v = \frac{(2s+1)^5}{10} \).

Step 4: Apply Integration by Parts Formula

Use the formula \( \int u \, dv = uv - \int v \, du \). Substitute \( u = s \), \( v = \frac{(2s+1)^5}{10} \), and \( du = ds \). The expression becomes \( s \cdot \frac{(2s+1)^5}{10} - \int \frac{(2s+1)^5}{10} \cdot ds \).

Step 5: Simplify and Solve the Remaining Integral

The remaining integral is \( \int \frac{(2s+1)^5}{10} \cdot ds \). Following a similar substitution as before, set \( w = 2s+1 \), which gives \( \int \frac{w^5}{20} \cdot dw = \frac{w^6}{120} + C \). Substitute back \( w = 2s+1 \) to get \( \frac{(2s+1)^6}{120} + C \).

Step 6: Combine the Solutions

Substitute the solutions into the integration by parts formula: \( \frac{s(2s+1)^5}{10} - \frac{(2s+1)^6}{120} + C \). Simplify the expression if necessary.

Key Concepts

Definite Integrals

Definite integrals are used to calculate the total accumulation of a quantity, often the area under a curve, between two points. They are represented as:

• \( \int_{a}^{b} f(x) \, dx \)

Here, \(a\) and \(b\) are the limits of integration and \(f(x)\) is the function being integrated. The result provides a numerical value that represents the net area under the curve from \(x = a\) to \(x = b\).
To solve a definite integral, one typically finds the indefinite integral (antiderivative) of the function first, and then uses the Fundamental Theorem of Calculus to evaluate it at the upper and lower limits. This means calculating the indefinite integral and then subtracting its value at \(a\) from its value at \(b\). If integration by parts or substitution is used, make sure to return to the original variable before evaluating at the limits.

Integral Calculus

Integral calculus is one of the two main branches of calculus, the other being differential calculus. It focuses on the concept of integration, which is the process of finding a function given its derivative. Integration can be used for a variety of applications, such as computing areas, volumes, central points, and the total accumulated quantities.
The basic idea of integration is finding the antiderivative, or indefinite integral, which involves determining a function whose derivative gives the original function. Rules for finding antiderivatives include reversing the process of differentiation, using techniques such as:

• Power Rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
• Basic Trigonometric: \( \int \cos(x) \, dx = \sin(x) + C \)

Methods like integration by parts and substitution (as highlighted in the given step-by-step solution) are more advanced techniques designed to handle more complex integrals which don't solve directly from basic rules.

U-Substitution

U-substitution, or substitution method, is a common technique in integral calculus used to simplify the process of finding the antiderivative of a function. This technique is particularly useful when dealing with composite functions.
The idea is to substitute part of the integral with a new variable \(u\), turning a complex integral into a simpler one:

• Choose \(u\) to simplify the integral, typically \( u = g(x) \), where \( g'(x) \, dx \) is present in the integral.
• Find \( du = g'(x) \, dx \).

This transforms the integral from \( \int f(g(x))g'(x) \, dx \) to \( \int f(u) \, du \).
In the original problem solution, a substitution \( w = 2s+1 \) was used, simplifying the integration step. By rewriting \( ds = \frac{1}{2} \, dw \), the integral \( \int (2s+1)^4 \, ds \) became easier to evaluate by turning it into a polynomial form in terms of \(w\). This method is effective in breaking down more tedious integrals, achieving a simpler expression that is easier to integrate.