Problem 3
Question
In Problems 3 through 8, nd the inde nite integrals. (a) \(\int(2 x+1)^{3} d x\) (b) \(\int \frac{1}{(2 x+1)^{2}} d x\) (c) \(\int \frac{1}{(2 x+1)} d x\) (d) \(\int \frac{1}{\sqrt{2 x+1}} d x\)
Step-by-Step Solution
Verified Answer
(a) The indefinite integral of \((2x+1)^3\) with respect to x is \(\frac{1}{8}(2x+1)^4 + C\) (b) The indefinite integral of \(\frac{1}{(2x+1)^2}\) with respect to x is \(\frac{1}{2(2x+1)} + C\) (c) The indefinite integral of \(\frac{1}{(2x+1)}\) with respect to x is \(\frac{1}{2} \ln |2x+1| +C\) (d) The indefinite integral of \(\frac{1}{\sqrt{2x+1}}\) with respect to x is \(\sqrt{2x+1} + C\)
1Step 1: Integral of (2x+1)^3
Let \( u = 2x + 1 \). Then \( du = 2dx \). So we need to express the integral in terms of \( u \). Thus, the integral becomes \(\frac{1}{2} \int u^3 du\), which equals \(\frac{1}{2} * \frac{1}{4}u^4 + C = \frac{1}{8}(2x+1)^4 + C\)
2Step 2: Integral of 1/(2x+1)^2
Again, let \( u = 2x + 1 \), hence \( du = 2dx \). The integral can now be rewritten as \( -\frac{1}{2} \int u^{-2} du\), which equals \(-\frac{1}{2} * -u^{-1} + C = \frac{1}{2(2x+1)} + C\)
3Step 3: Integral of 1/(2x+1)
With \( u = 2x + 1 \), and \( du = 2dx \), integrate \(\frac{1}{2} \int u^{-1} du\), which equals \( \frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |2x+1| +C\)
4Step 4: Integral of 1/sqrt(2x+1)
We substitute \( u = 2x + 1 \), making \( du = 2 dx\), and the integral simplifies to \(\frac{1}{2} \int u^{-1/2} du\). This equals \(\frac{1}{2} * 2u^{1/2} + C = \sqrt{2x+1} + C\)
Key Concepts
Integration TechniquesSubstitution MethodIntegral CalculusAntiderivatives
Integration Techniques
Mastering integration techniques is a vital skill within the study of integral calculus. These include methods for transforming complex integrals into simpler forms to facilitate their evaluation.
One standard technique, showcased in the exercises from the textbook, involves algebraic manipulation to simplify the integrand before integration. For example, knowing that the integral of a sum of functions is the sum of the integrals of those functions can guide students in approaching \textbf{Exercise (a)}, where the integral of \( (2x+1)^3 \) can be viewed as the sum of the integrals of its expanded form.
Another crucial technique is the \textbf{reverse power rule}, which applies to \textbf{Exercise (b)}, where integrating \( u^n \) yields \( \frac{u^{n+1}}{n+1} \) when \( n \) is not equal to \( -1 \). This rule is a cornerstone of integrating polynomials and their reciprocal powers.
One standard technique, showcased in the exercises from the textbook, involves algebraic manipulation to simplify the integrand before integration. For example, knowing that the integral of a sum of functions is the sum of the integrals of those functions can guide students in approaching \textbf{Exercise (a)}, where the integral of \( (2x+1)^3 \) can be viewed as the sum of the integrals of its expanded form.
Another crucial technique is the \textbf{reverse power rule}, which applies to \textbf{Exercise (b)}, where integrating \( u^n \) yields \( \frac{u^{n+1}}{n+1} \) when \( n \) is not equal to \( -1 \). This rule is a cornerstone of integrating polynomials and their reciprocal powers.
Substitution Method
The substitution method, often referred to as u-substitution, is where we choose a part of the integrand to be \( u \) and replace the rest with \( du \). This 'substitution' effectively changes the integral into a simpler form, usually a basic antiderivative that is easier to solve.
For instance, in the textbook exercises, each problem effectively utilizes substitution. Examine \textbf{Exercise (d)}: By setting \( u = 2x + 1 \) and substituting \( dx \) for \( \frac{1}{2}du \), the integral of \( \frac{1}{\sqrt{2x+1}} \) becomes a straightforward power function in terms of \( u \) that can be integrated directly. Students should note that finding the right substitution is a crucial skill that often requires practice and insight into the structure of the function to be integrated.
For instance, in the textbook exercises, each problem effectively utilizes substitution. Examine \textbf{Exercise (d)}: By setting \( u = 2x + 1 \) and substituting \( dx \) for \( \frac{1}{2}du \), the integral of \( \frac{1}{\sqrt{2x+1}} \) becomes a straightforward power function in terms of \( u \) that can be integrated directly. Students should note that finding the right substitution is a crucial skill that often requires practice and insight into the structure of the function to be integrated.
Integral Calculus
Integral calculus is one of the two primary branches of calculus, the other being differential calculus. It primarily deals with finding the quantity where the rate of change is known. This concept helps us understand the accumulation of quantities, such as area under a curve, volume, and other concepts that stem from a sum of infinitesimally small amounts.
In the scope of the provided exercises, integral calculus allows us to determine the antiderivative of algebraic and transcendental functions. Each subproblem highlights a different integral calculus application, for example, using the properties of logarithms to integrate a function in the form of \( \frac{1} {ax+b} \) as seen in \textbf{Exercise (c)}. The outcome of these problems often includes the +C term, representing the constant of integration. This term is crucial because the process of integration, unlike differentiation, does not yield a unique function.
In the scope of the provided exercises, integral calculus allows us to determine the antiderivative of algebraic and transcendental functions. Each subproblem highlights a different integral calculus application, for example, using the properties of logarithms to integrate a function in the form of \( \frac{1} {ax+b} \) as seen in \textbf{Exercise (c)}. The outcome of these problems often includes the +C term, representing the constant of integration. This term is crucial because the process of integration, unlike differentiation, does not yield a unique function.
Antiderivatives
The concept of antiderivatives refers to finding a function whose derivative is the given function. In essence, it involves reversing the process of differentiation. Antiderivatives are the backbone of computing indefinite integrals, as seen in the provided textbook exercises.
Each integral in exercises (a) through (d) seeks the antiderivative of a function. For example, the solution to \textbf{Exercise (a)} gives us \( \frac{1}{8}(2x+1)^4 + C \), which means that the derivative of \( \frac{1}{8}(2x+1)^4 \) with respect to \( x \) is \( (2x+1)^3 \). The constant \( C \) is included because the derivative of any constant is zero, and therefore, it would not appear in the original function. For any integral calculus problem, finding antiderivatives is crucial since it informs us about the accumulation of quantities, such as the area under the curve.
Each integral in exercises (a) through (d) seeks the antiderivative of a function. For example, the solution to \textbf{Exercise (a)} gives us \( \frac{1}{8}(2x+1)^4 + C \), which means that the derivative of \( \frac{1}{8}(2x+1)^4 \) with respect to \( x \) is \( (2x+1)^3 \). The constant \( C \) is included because the derivative of any constant is zero, and therefore, it would not appear in the original function. For any integral calculus problem, finding antiderivatives is crucial since it informs us about the accumulation of quantities, such as the area under the curve.
Other exercises in this chapter
Problem 2
Compute the integral. \(\int A t^{n} d t\), where \(A\) and \(n\) are constants and \(n \neq-1 .\)
View solution Problem 3
Find the given indefinite integral. $$ \int x \sqrt{3 x+5} d x $$
View solution Problem 3
Compute the integral. \(\int 3 x^{-1} d x\)
View solution Problem 4
Find the given indefinite integral. $$ \frac{2 x}{3+x} d x $$
View solution