/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 Evaluate each definite integral ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate each definite integral using integration by parts. (Leave answers in exact form.) \(\int_{0}^{2} x e^{x} d x\)

Short Answer

Expert verified
The integral evaluates to \(e^{2} - 1\).

Step by step solution

01

Choose Functions for Integration by Parts

Integration by parts formula is \( \int u \mathrm{d}v = uv - \int v \mathrm{d}u \). Choose \( u = x \) and \( \mathrm{d}v = e^x \mathrm{d}x \).
02

Differentiate and Integrate Chosen Functions

Differentiating \( u = x \) gives \( \mathrm{d}u = \mathrm{d}x \). Integrating \( \mathrm{d}v = e^x \mathrm{d}x \) gives \( v = e^x \).
03

Apply Integration by Parts Formula

Substitute the chosen functions into the integration by parts formula: \( \int x e^{x} \mathrm{d}x = x e^{x} - \int e^{x} \mathrm{d}x \).
04

Evaluate the Remaining Integral

The remaining integral is \( \int e^{x} \mathrm{d}x = e^x + C \). Thus, \( \int x e^{x} \mathrm{d}x = x e^{x} - e^{x} + C \).
05

Apply Limits of Definite Integral

Evaluate \([0, 2]\) for \( x e^{x} - e^{x} \). This becomes \([2e^{2} - e^{2}] - [0e^{0} - e^{0}] = e^{2} - 1\).

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Key Concepts

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

Definite Integrals
Definite integrals calculate the area under a curve over a specific interval. They have clear boundaries, known as limits of integration, which indicate where to start and stop. For the exercise at hand, this interval is from 0 to 2, denoted as \(\int_{0}^{2}\).
When you evaluate a definite integral, you determine the net area between the curve and the x-axis, considering parts above the x-axis as positive and below as negative. The process results in a specific numerical value, which represents the accumulated value over the given interval. In this problem, by applying the definite integral over the function \(x e^{x}\), the solution should yield an exact number, providing insight into the behavior of the function across the interval from 0 to 2.
  • Choose the function and limits of integration carefully.
  • Apply limits after performing integration.
  • Definite integrals always result in a finite number.
Integration Techniques
There are various integration techniques in calculus to find the integral of different functions. One such technique is **Integration by Parts**, useful when integrating products of functions. This involves breaking down a given integral into parts that are simpler to evaluate.
Integration by parts leverages the formula \(\int u \mathrm{d}v = uv - \int v \mathrm{d}u\), transforming an often difficult integral into a more manageable form. Choosing the appropriate functions \(u\) and \(\mathrm{d}v\) is crucial. An ideal approach is to select a \(u\) that simplifies upon differentiation and a \(\mathrm{d}v\) that is straightforward to integrate. In the given problem:
  • **Chosen functions**: \(u = x\) and \(\mathrm{d}v = e^x \mathrm{d}x\).
  • **Differentiation and Integration**: Differentiate \(u\) to get \(\mathrm{d}u\) and integrate \(\mathrm{d}v\) to find \(v\).
  • **Apply the formula**: Substitute and simplify the integral.
Calculus Problems
Calculus problems often demand a blend of integration techniques and a strategic approach. When faced with problems that involve definite integrals, like \(\int_{0}^{2} x e^{x} \mathrm{d}x\), it is essential to:
- Break down the problem step by step.- Select the correct method, here it is the integration by parts.- Follow through with persistence and attention to detail.
In our scenario, the calculus problem required finding an integral for a product of polynomial and exponential functions over a set interval. It necessitated applying integration by parts to simplify and compute the integral efficiently. Once simplified, evaluate at the prescribed limits:
  • Calculate the function's result at the upper limit.
  • Subtract the function's result at the lower limit from the above.
  • The difference is the solution, providing an exact form \(e^{2} - 1\).

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Most popular questions from this chapter

73-74. GENERAL: Permanent Endowments The formula for integrating the exponential function \(a^{b x}\) is \(\int a^{b x} d x=\frac{1}{b \ln a} a^{b x}+C\) for constants \(a>0\) and \(b,\) as may be verified by using the differentiation formulas on page 289. Use the formula above to find the size of the permanent endowment needed to generate an annual \(\$ 12,000\) forever at \(6 \%\) interest compounded annually. [Hint: Find \(\left.\int_{0}^{\infty} 12,000 \cdot 1.06^{-x} d x .\right]\) Compare your answer with that found in Exercise 43 (page 413 ) for the same interest rate but compounded continuously.

BUSINESS: Sales A publisher estimates that a book will sell at the rate of \(16,000 e^{-0.8 t}\) books per year \(t\) years from now. Find the total number of books that will be sold by summing (integrating) this rate from 0 to \(\infty\).

Repeated Integration by Parts Using a Table The solution to a repeated integration by parts problem can be organized in a table. As an example, we solve \(\int x^{2} e^{3 x} d x .\) We begin by choosing $$ u=x^{2} \quad d v=v^{\prime} d x=e^{3 \tau} d x $$ We then make a table consisting of the following three columns: Finally, the solution is found by adding the signed products of the diagonals shown in the table: $$ \int x^{2} e^{3 x} d x=\frac{1}{3} x^{2} e^{3 x}-\frac{2}{9} x e^{3 x}+\frac{2}{27} e^{3 x}+C $$ After reading the preceding explanation, find each integral by repeated integration by parts using a table. \(\int x^{2} e^{2 x} d x\)

17-40. Evaluate each improper integral or state that it is divergent. $$ \int_{-\infty}^{0} \frac{1}{1-x} d x $$

Sometimes an integral requires two or more integrations by parts. As an example, we apply integration by parts to the integral \(\int x^{2} e^{x} d x\). \(\int \underbrace{x^{2} e^{x} d x}=x^{2} e^{x}-\int_{u} \underbrace{e^{x} 2 x d x}=x^{2} e^{x}-2 \int x e^{x} d x\) \(\left[\begin{array}{cc}u=x^{2} & d v=e^{x} d x \\ d u=2 x d x & v=\int e^{x} d x=e^{x}\end{array}\right]\) The new integral \(\int x e^{x} d x\) is solved by a second integration by parts. Continuing with the previous solution, we choose new \(u\) and \(d u\) : \(=x^{2} e^{x}-2\left(\int x e^{x} d x\right) \quad\left[\begin{array}{c}u=x \quad d v=e^{x} d x \\ d u=d x \quad v=e^{x}\end{array}\right]\) \(=x^{2} e^{x}-2\left(x e^{x}-\int e^{x} d x\right)\) \(=x^{2} e^{x}-2\left(x e^{x}-e^{x}\right)+C\) \(=x^{2} e^{x}-2 x e^{x}+2 e^{x}+C\) After reading the preceding explanation, find each integral by repeated integration by parts. \(\int(x+1)^{2} e^{x} d x\)

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