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Chapter 7: Problem 34

Evaluate the following definite integrals. $$\int_{0}^{\ln 2} x e^{x} d x$$

Short Answer

Expert verified
Question: Evaluate the definite integral: $$\int_{0}^{\ln 2} x e^{x} d x$$ Answer: $$2 \ln 2 - 1$$

Step by step solution

01

Choose u and dv

Since we want to simplify the integrand and, integration by parts requires differentiation of one function and integration of the other, we choose: $$ u = x \quad \text{and} \quad d v = e^{x} d x $$
02

Differentiate u and integrate dv

Finding du and v: $$ d u =d x \quad \text{and} \quad v =\int e^{x} d x=e^{x} $$
03

Apply integration by parts

Integration by parts states that: $$ \int u d v = u v-\int v d u $$ Plugging in the values we found, $$ \int x e^{x} d x = x e^{x} - \int e^{x} d x $$
04

Integrate the remaining part

Now we integrate the remaining part: $$ \int e^{x} d x = e^{x} $$ So the integral of the integrand is: $$ \int x e^{x} d x=x e^{x}-e^{x}+C $$
05

Apply the limits and calculate the definite integral

Using the limits \(0\) and \(\ln 2\): $$ \int_{0}^{\ln 2} x e^{x} d x = \left[ x e^{x} - e^{x} \right]_{0}^{\ln 2} $$ Evaluating the antiderivative at the limits, we obtain: $$ \left[ (\ln 2) e^{\ln 2} - e^{\ln 2} - (0 \cdot e^{0}-e^{0}) \right] $$ Simplify the expression: $$ \left[ 2 \ln 2 - 2 - (0 - 1)\right] = 2 \ln 2-1 $$ So the value of the given definite integral is: $$ \int_{0}^{\ln 2} x e^{x} d x = 2 \ln 2-1 $$

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Key Concepts

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

Definite Integral
In calculus, a definite integral is a fundamental concept that represents the area under the curve of a function, between two specific points. It's not just a number; it has a geometric interpretation. When we evaluate a definite integral, we find the net area between the function and the x-axis from one point to another.
  • The integral symbol \( \int \) denotes the process of integration.
  • The limits of integration provide the specific bounds, in our case \(0\) and \(\ln 2\).
  • The function \(x e^x\) is the integrand, which is the function to be integrated.
Using definite integrals, we can measure quantities such as area, displacement, and in more general terms, accumulation of quantities. We use definite integrals in real-life scenarios like physics for displacement and work done by a force, which makes learning this concept extremely valuable.
Exponential Function
An exponential function is a mathematical expression where the variable appears as an exponent. In simpler terms, it involves expressions like \( e^x \), where \( e \) is the base of the natural logarithm, approximately equal to 2.718.
  • This base \( e \) is irrational and serves as the foundation for natural exponential functions.
  • In the given exercise, \( e^x \) was multiplied by \( x \), making the integrand \( x e^x \) a product of a linear and an exponential function.
  • Exponential functions grow quickly鈥攕ometimes describing real-world phenomena like population growth and radioactive decay.
These functions are crucial for integration problems as they often appear in calculus equations due to their distinct growth rate properties. Understanding how to manipulate and integrate exponential functions is a cornerstone of calculus.
Antiderivative
An antiderivative, also known as an indefinite integral, is a function whose derivative is the original function. Essentially, it's the reverse process of differentiation.
  • Finding the antiderivative is crucial in solving integration problems like the one in our exercise.
  • For \( e^x \), the antiderivative is straightforwardly \( e^x \) because the derivative of \( e^x \) is also \( e^x \).
  • In our definite integral task, once we applied integration by parts, exploring the antiderivative of remaining functions was key to evaluating the integral.
The notation \( \int \) indicates an antiderivative and represents the general form of integration. Meanwhile, the constant \( C \) in an indefinite integral accounts for all possible vertical shifts. Understanding antiderivatives prepares students to deal with more complex integration techniques like integration by parts.
Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It serves as a framework for understanding change and motion, often applied in diverse fields such as physics, engineering, and economics.
  • It provides tools to measure things that change, like growth rates and dynamic systems.
  • The two main branches are differential calculus (concerning rates of change) and integral calculus (concerned with accumulation), which intertwine beautifully.
  • In the exercise, integration by parts鈥攁 technique rooted in integral calculus鈥攚as used to solve the problem efficiently.
Mastering calculus concepts enables analytical problem-solving skills critical for higher education and beyond. As you delve into problems like evaluating definite integrals, you tap into calculus's power to illuminate complex changes in processes and systems.

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Most popular questions from this chapter

Evaluating an integral without the Fundamental Theorem of Calculus Evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) using the following steps. a. If \(f\) is integrable on \([0, b],\) use substitution to show that $$ \int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x $$ b. Use part (a) and the identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) (Source: The College Mathematics Journal 33, 4, Sep 2004).

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 2} \frac{d \theta}{\cos \theta+\sin \theta}$$

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{-|x|} d x\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

Compare the errors in the Midpoint and Trapezoid Rules with \(n=4,8,16,\) and 32 subintervals when they are applied to the following integrals (with their exact values given). \(\int_{0}^{1}\left(8 x^{7}-7 x^{8}\right) d x=\frac{2}{9}\)

The reaction of chemical compounds can often be modeled by differential equations. Let \(y(t)\) be the concentration of a substance in reaction for \(t \geq 0\) (typical units of \(y\) are moles/L). The change in the concentration of the substance, under appropriate conditions, is \(\frac{d y}{d t}=-k y^{n},\) where \(k>0\) is a rate constant and the positive integer \(n\) is the order of the reaction. a. Show that for a first-order reaction \((n=1),\) the concentration obeys an exponential decay law. b. Solve the initial value problem for a second-order reaction \((n=2)\) assuming \(y(0)=y_{0}\). c. Graph and compare the concentration for a first-order and second-order reaction with \(k=0.1\) and \(y_{0}=1\).
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