Problem 9

Question

Solve using integration by parts, $\int u(x) v^{\prime}(x) d x=u(x) v(x)-\int v(x) u^{\prime}(x) d x$ (or $\left.\int u d v=u v-\int v d u\right)$ a. $\int x e^{x} d x$ b. $\int x \ln x d x$ c. $\int x \sin x d x$ d. $\int x^{2} e^{x} d x$ e. $\int x e^{2 x} d x$ f. $\int \ln x \cdot 1 d x$ g. $\int x \cos x d x$ h. $\int x^{3} e^{x^{2}} d x$

Step-by-Step Solution

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Answer

Each integral uses integration by parts in various ways, resulting in simplified expressions.

1Step 1: Understanding the Problem

We will use the formula $ \int u \, dv = uv - \int v \, du $ to solve a series of integrals. This approach is known as integration by parts. It is especially useful for integrating products of functions.

2Step 2: Part a: Choosing Functions for Integration by Parts

For $ \int x e^x \, dx $, choose $ u = x $ and $ dv = e^x \, dx $. Differentiating, we get $ du = dx $, and integrating $ dv $ gives $ v = e^x $.

3Step 3: Part a: Applying Integration by Parts

Using the formula $ \int u \, dv = uv - \int v \, du $, substitute: \[ \int x e^x \, dx = x e^x - \int e^x \, dx \]. This simplifies to $ x e^x - e^x + C $.

4Step 4: Part b: Choosing Functions for Integration by Parts

For $ \int x \ln x \, dx $, choose $ u = \ln x $ and $ dv = x \, dx $. Then, $ du = \frac{1}{x} \, dx $ and $ v = \frac{x^2}{2} $.

5Step 5: Part b: Applying Integration by Parts

Substitute into the formula: \[ \int x \ln x \, dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \]. This simplifies to $ \frac{x^2}{2} \ln x - \frac{x^2}{4} + C $.

6Step 6: Part c: Choosing Functions for Integration by Parts

For $ \int x \sin x \, dx $, let $ u = x $ and $ dv = \sin x \, dx $. Then, $ du = dx $ and integrating $ dv $, we get $ v = -\cos x $.

7Step 7: Part c: Applying Integration by Parts

The integral becomes \[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \]. This simplifies to $ -x \cos x + \sin x + C $.

8Step 8: Part d: Choosing Functions for Integration by Parts

For $ \int x^2 e^x \, dx $, choose $ u = x^2 $ and $ dv = e^x \, dx $. Then, $ du = 2x \, dx $ and $ v = e^x $.

9Step 9: Part d: Applying Integration by Parts

Apply integration by parts twice: \[ x^2 e^x - \int 2x e^x \, dx \]. Re-apply integration by parts to $ \int 2x e^x \, dx $ to find the complete solution as $ (x^2 - 2x + 2) e^x + C $.

10Step 10: Part e: Choosing Functions for Integration by Parts

For $ \int x e^{2x} \, dx $, let $ u = x $ and $ dv = e^{2x} \, dx $. Thus, $ du = dx $ and $ v = \frac{1}{2} e^{2x} $.

11Step 11: Part e: Applying Integration by Parts

The integral becomes: \[ x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx \], simplifying to $ \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C $.

12Step 12: Part f: Choosing Functions for Integration by Parts

For $ \int \ln x \cdot 1 \, dx $, choose $ u = \ln x $ and $ dv = dx $. Thus, $ du = \frac{1}{x} \, dx $ and $ v = x $.

13Step 13: Part f: Applying Integration by Parts

Substituting gives \[ x \ln x - \int x \cdot \frac{1}{x} \, dx \], which reduces to $ x \ln x - x + C $.

14Step 14: Part g: Choosing Functions for Integration by Parts

For $ \int x \cos x \, dx $, take $ u = x $ and $ dv = \cos x \, dx $. Thus, $ du = dx $ and $ v = \sin x $.

15Step 15: Part g: Applying Integration by Parts

The integral becomes \[ x \sin x - \int \sin x \, dx \], which simplifies to $ x \sin x + \cos x + C $.

16Step 16: Part h: Choosing Functions for Integration by Parts

For $ \int x^3 e^{x^2} \, dx $, set $ u = x^2 $ and $ dv = x e^{x^2} \, dx $. Therefore, $ du = 2x \, dx $ and $ v = \frac{1}{2} e^{x^2} $.

17Step 17: Part h: Applying Integration by Parts

Substitute to find \[ \frac{1}{2} x^2 e^{x^2} - \int \frac{1}{2} \cdot 2x e^{x^2} \, dx \], which simplifies directly to $ \frac{1}{2} e^{x^2} + C $.

Key Concepts

Integration TechniquesCalculus Problem-SolvingAdvanced Calculus Concepts

Integration Techniques

Integration is a key concept in calculus that connects the area under a curve with antiderivatives. Among various integration techniques, "integration by parts" is particularly important for dealing with products of functions. It stems from the product rule for differentiation and is expressed in the formula: $ \int u \, dv = uv - \int v \, du $. This formula helps transform a complex integral into a potentially simpler one.

When choosing the

function $ u $ — typically pick the function that simplifies upon differentiation
function $ dv $ — choose the one that simplifies when integrated

For example, in solving $ \int x e^x \, dx $:

Set $ u = x $ (since its derivative simplifies to 1)
Set $ dv = e^x \, dx $ (because integrating $ e^x $ is straightforward)

The strategic choice of these functions is crucial for solving integrals effectively. Remember, integration by parts can often be iterative, requiring multiple applications for complex expressions.

Calculus Problem-Solving

At its core, calculus is a tool for problem-solving, especially when dealing with dynamic systems. Integration by parts is a key problem-solving technique in calculus that helps tackle integrals that involve products of algebraic and transcendental functions like polynomials and logarithmic functions.

To solve integration problems using this technique, one must:

Identify the type of functions in the integral
Apply the integration by parts formula thoughtfully
Reapply the method if the resultant integral remains complex

For example, when integrating $ \int x \ln x \, dx $, the following approach works:

Choose $ u = \ln x $ to simplify upon differentiation, making $ du = \frac{1}{x} \, dx $
Choose $ dv = x \, dx $ which integrates to $ v = \frac{x^2}{2} $

Failing to choose wisely could lead to more daunting integrals. Sometimes, when performing calculus problem-solving, an intuition for function behavior and transformations is valuable.

Advanced Calculus Concepts

Delving into advanced calculus involves confronting integrals that are not straightforward. Integration by parts is among the advanced techniques that unpack challenging integrals, particularly those involving exponential functions, like $ e^x $, or trigonometric functions, such as $ \sin x $ or $ \cos x $.

Some tips for handling advanced calculus concepts include:

Recognizing patterns — noticing similarities in various integrals can streamline the problem-solving process
Breaking down complex integrals into manageable parts
Maintaining flexibility — some integrals may require rethinking the approach and trying different functions for $ u $ and $ dv $

Take solving $ \int x^2 e^x \, dx $ as an example:

Initially choose $ u = x^2 $ and $ dv = e^x \, dx $
Simplifying further requires a recursive application of integration by parts
Eventually yielding $ (x^2 - 2x + 2) e^x + C $

This illustrates how layered applications of integration by parts unravel intricate calculus problems effectively.

Problem 8

Problem 10

Other exercises in this chapter

Problem 7

Let $F(t)=[t]$ where $[t]$ is the greatest integer less than or equal to $t$. For example, $[\pi]=3,$ and $[2]=2 .$ Let $G(x)=\int_{0}^{x} f(t) d t$

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Problem 8

a. (a) Compute $[x \ln x-x]^{\prime}$ (b) Compute $\int_{1}^{e} \ln x d x$ b. Note: $\sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}$ Compute \(\int_{0}^{\pi}

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Problem 10

Exercise 10.5 .10 a. Use integration by parts on $\int e^{x} \sin x d x,$ with $u(x)=e^{x}$ and $v^{\prime}(x)=\sin x,$ to show that $$ \int e^{x} \sin x

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Problem 7

Compute the following antiderivatives. If you solve the integral by a substitution, $u(t)=$, then identify in writing $u(t)$ and $u^{\prime}(t)$. a. \(\in

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