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Chapter 10: Problem 9

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. \(\int x^{3} e^{x^{2}} d x\)

Short Answer

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

Step by step solution

01

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.
02

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 \( v = e^x \).
03

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 \( x e^x - e^x + C \).
04

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 \( v = \frac{x^2}{2} \).
05

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 \( \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \).
06

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 \( v = -\cos x \).
07

Part c: Applying Integration by Parts

The integral becomes \[ \int x \sin x \, dx = -x \cos x + \int \cos x \, dx \]. This simplifies to \( -x \cos x + \sin x + C \).
08

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 \( v = e^x \).
09

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 \( (x^2 - 2x + 2) e^x + C \).
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 \( v = \frac{1}{2} e^{2x} \).
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 \( \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x} + C \).
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 \( v = x \).
13

Part f: Applying Integration by Parts

Substituting gives \[ x \ln x - \int x \cdot \frac{1}{x} \, dx \], which reduces to \( x \ln x - x + C \).
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 \( v = \sin x \).
15

Part g: Applying Integration by Parts

The integral becomes \[ x \sin x - \int \sin x \, dx \], which simplifies to \( x \sin x + \cos x + C \).
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 \( v = \frac{1}{2} e^{x^2} \).
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 \( \frac{1}{2} e^{x^2} + C \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 \( u \) 鈥� typically pick the function that simplifies upon differentiation
  • function \( dv \) 鈥� 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 鈥� noticing similarities in various integrals can streamline the problem-solving process
  • Breaking down complex integrals into manageable parts
  • Maintaining flexibility 鈥� 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.

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Most popular questions from this chapter

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 \(\int_{0}^{\pi} \sin ^{2} \theta d \theta\) c. (a) Show that \(\frac{1}{t}+\frac{1}{1-t}=\frac{1}{t(1-t)}\) (b) Compute \(\int \frac{1}{t(1-t)} d t\) d. \(\quad\) (a) Believe: \([\arcsin x]^{\prime}=\frac{1}{\sqrt{1-x^{2}}}\) (b) Compute \(\int_{0}^{1 / 2} \frac{1}{\sqrt{1-x^{2}}} d x\)

Evaluate the integrals. a. \(\int_{1}^{3} t^{2} d t\) b. \(\int_{0}^{2} t^{3} d t\) c. \(\int_{0}^{2} e^{t} d t\) d. \(\int_{1}^{3} \frac{1}{t} d t\) e. \(\int_{1}^{3}(1+t)^{2} d t\) f. \(\int_{1}^{3} 5 t d t\) g. \(\int_{0}^{2} t+5 d t\) $$ \text { h. } \int_{0}^{2} e^{2 t} d t \quad \text { i. } \quad \int_{0}^{3} e^{-t} d t \quad \text { j. } \quad \int_{1}^{3}\left(1+t^{2}\right) d t $$

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 $$ \int e^{x} \sin x d x=-e^{x} \cos x+\int e^{x} \cos x d x $$ Use a second step integration by parts on \(\int e^{x} \cos x d x\) to show that $$ \int e^{x} \cos x d x=e^{x} \sin x-\int e^{x} \sin x d x $$ Combine the previous two equations to show that $$ \int e^{x} \sin x d x=\frac{1}{2} e^{x}(\sin x-\cos x)+C $$ b. Do two steps of integration by parts on \(\int e^{x} \cos x d x \quad\) and show that \(\quad \int e^{x} \cos x d x=\frac{1}{2} e^{x}(\sin x+\cos x)+C\) c. Do two steps of integration by parts on \(\int(\sin x) e^{-x} d x \quad\) and show that \(\quad \int_{0}^{\pi} e^{-x} \sin x d x=\frac{1}{2}\left(e^{-\pi}+1\right)\) d. Clever! Note that \(\int e^{\sqrt{x}} d x=\int 2 \sqrt{x} e^{\sqrt{x}} \frac{1}{2 \sqrt{x}} d x\). Let \(u=2 \sqrt{x}\) and \(v^{\prime}=e^{\sqrt{x}} \frac{1}{2 \sqrt{x}}\) and show that $$ \int e^{\sqrt{x}} d x=2 \sqrt{x} e^{\sqrt{x}}-2 e^{\sqrt{x}}+C $$

Let \(F(t)=[t]\) where \([t]\) is the greatest integer less than or equal to \(t\). For example, \([\pi]=3,\) and \([2]=2 .\) Let \(G(x)=\int_{0}^{x} f(t) d t\) for \(0 \leq x \leq 4\) a. Sketch the graph of \(f\) for \(0 \leq t \leq 5\) b. Sketch the graph of \(G\). c. Sketch the graph of \(G^{\prime}\). d. You should find that \(G^{\prime} \neq f .\) Does this contradict the Fundamental Theorem of Calculus?

At 1: 00 a.m. an oil pipe line bursts and starts releasing oil into a lake at the rate of 2 cubic meters per hour. At 2: 00 a.m., a second oil pipe line bursts and also starts releasing oil into the lake at the rate of 3 cubic meters per hour. a. How much oil is in the lake at 1: 00,1: 30,2: 00,2: 30,3: 00,\(3: 30,\) and \(4: 00 ?\) b. Let \(\mathrm{T}(\mathrm{x})\) be the total amount of oil in the lake at time \(x\). Draw a graph of \(T\). c. Write equations describing the total amount of oil, \(T(x),\) in the lake for each time \(x\) between 1:00 a.m. and 4: 00 a.m. d. Compute \(T^{\prime}\).
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