Problem 16
Question
Find the indefinite integral and check the result by differentiation. $$ \int x\left(1-2 x^{2}\right)^{3} d $$
Step-by-Step Solution
Verified Answer
The indefinite integral of \(x(1-2x^2)^3\) is \(-\frac{1}{16}(1-2x^2)^4\).
1Step 1: Identify the Integral to Solve
The task is to find the indefinite integral of \(x(1-2x^2)^3\). The integral to solve is: \(\int x(1-2x^2)^3 dx\).
2Step 2: Use Substitution for Simplification
Let \(u=1 - 2 x^2\). Now differentiating 'u' with respect to 'x' gives: \(du = -4x dx\). In order to make our integral have this form, we multiply and divide our integral by -4. We get: \(-\frac{1}{4} \int u^3(-4x dx)\). Now subtitute \(du\) for \(-4x dx\), We have: \(-\frac{1}{4} \int u^3 du\).
3Step 3: Solve the Simplified Integral
Now solve the integral by using the power rule of integration, which states that, the integral of \(x^n\) with respect to 'x' is \(\frac{x^{n+1}}{n+1}\). Here n=3. So, \(\int u^3 du = \frac{u^4}{4}\) . Substituting back, we have \(-\frac{1}{4} * \frac{u^4}{4} = -\frac{1}{16}u^4\). Re-substituting for \(u\), we have the integral as \(-\frac{1}{16}(1-2x^2)^4\)
4Step 4: Checking the result by Differentiation
To check our result, we differentiate \(-\frac{1}{16}(1-2x^2)^4\) with respect to 'x'. Following the Chain Rule for differentiation and simplifying the result, we see that this gives us back \(x(1-2x^2)^3\), indicating that our original integration was performed correctly.
Key Concepts
Substitution MethodIntegration by PartsPower Rule for Integration
Substitution Method
The substitution method in integration is a powerful technique used to transform complex integrals into simpler ones. It resembles the method of substitution in algebra, where you set variables equal to each other to simplify expressions. In this context, our goal is to change variables to make the integral easier to manage.
To illustrate, consider the integral \( \int x(1-2x^2)^3 \, dx \). The expression inside the integral looks complex, mainly because of the \( (1-2x^2)^3 \) term. So, we use substitution. We let \( u = 1 - 2x^2 \).
This substitution helps by leaving us with a simpler integral in terms of \( u \). Differentiating the new variable, we find \( du = -4x \, dx \).
To illustrate, consider the integral \( \int x(1-2x^2)^3 \, dx \). The expression inside the integral looks complex, mainly because of the \( (1-2x^2)^3 \) term. So, we use substitution. We let \( u = 1 - 2x^2 \).
This substitution helps by leaving us with a simpler integral in terms of \( u \). Differentiating the new variable, we find \( du = -4x \, dx \).
- We rearrange to get \(dx\) and \(x\) terms in line with \( du \), resulting in \(-\frac{1}{4} \int u^3 (-4x \, dx) \).
- Then, we substitute \( du \) for \(-4x \, dx\), leading to the integral \(-\frac{1}{4} \int u^3 \, du \).
Integration by Parts
Integration by parts is another critical tool in integral calculus, especially useful when dealing with products of functions. It's based on the product rule from differentiation.
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here, \( u \) and \( dv \) are chosen such that the integral of \( v \, du \) is simpler than the original integral.
In the context of our problem \( \int x(1-2x^2)^3 \, dx \), integration by parts isn't directly applied as substitution provides a more straightforward path. However, understanding integration by parts is essential for integrating functions where substitution is unhelpful. To apply it, you'd generally:
The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \). Here, \( u \) and \( dv \) are chosen such that the integral of \( v \, du \) is simpler than the original integral.
In the context of our problem \( \int x(1-2x^2)^3 \, dx \), integration by parts isn't directly applied as substitution provides a more straightforward path. However, understanding integration by parts is essential for integrating functions where substitution is unhelpful. To apply it, you'd generally:
- Select \( u \) and \( dv \) appropriately from the original integral.
- Derive \( du \) by differentiating \( u \).
- Integrate \( dv \) to find \( v \).
- Substitute into the formula to simplify the expression.
Power Rule for Integration
The power rule for integration is one of the foundational rules that simplifies the integration process, especially when dealing with polynomials. It states that the integral of \( x^n \), where \( n eq -1 \), is given by \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), with \( C \) representing the constant of integration.
In the problem \( \int x(1-2x^2)^3 \, dx \), after substituting \( u = 1 - 2x^2 \), the integral became \(-\frac{1}{4} \int u^3 \, du \). This was a straightforward application of the power rule.
In the problem \( \int x(1-2x^2)^3 \, dx \), after substituting \( u = 1 - 2x^2 \), the integral became \(-\frac{1}{4} \int u^3 \, du \). This was a straightforward application of the power rule.
- We integrated \( u^3 \) to get \( \frac{u^4}{4} \).
- After computing the integral, we adjusted by multiplying with the out-front factor of \(-\frac{1}{4} \), yielding \(-\frac{1}{16}u^4 \).
- Finally, we substituted back our original expression for \( u \) to complete the solution.
Other exercises in this chapter
Problem 16
Sketch the region bounded by the graphs of the functions and find the area of the region. $$ y=x^{3}-2 x+1, y=-2 x, x=1 $$
View solution Problem 16
Use the Log Rule to find the indefinite integral. $$ \int \frac{1}{6 x-5} d x $$
View solution Problem 16
Find the indefinite integral and check your result by differentiation. $$ \int d r $$
View solution Problem 17
Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(x)=\sqrt[3]{x}, g(x)=x $$
View solution