/*! 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 35 Evaluate the definite integral. ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 35

Evaluate the definite integral. $$\int_{0}^{2} \frac{e^{x}}{1+e^{2 x}} d x$$

Short Answer

Expert verified
The definite integral from 0 to 2 of the given function is equal to: \( \arctan(e^2) - \frac{\pi}{4} \).

Step by step solution

01

Simplify the integrand

Firstly, let's set \( u = e^x \). By taking derivative both sides we have: \( du = e^x dx \). Now, we can rewrite our integral in terms of u, which is: \(\int_{1}^{e^2} \frac{u}{1+u^2} du\) where we have also changed our limits of integration by substituting x for u.
02

Perform the integration

Looking at the simplified integral, it should be clear that it's a standard integral and recognised as the derivative of the inverse tangent function. Thus, the integral simplifies to \( \arctan(u)\). So, the integral from 1 to \( e^2 \) of our simplified function is: \( \arctan(e^2) - \arctan(1) \).
03

Substituting the values

\(\arctan(1)\) is \( \frac{\pi}{4} \). Therefore, we have: \( \arctan(e^2) - \frac{\pi}{4} \). Now as \( e^2 > 1 \), \( \arctan(e^2) \) is between \( \frac{\pi}{2} \) and \( \frac{\pi}{4} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Substitution Method
The substitution method is a common technique used in calculus, especially when working with integrals. It's like a tool to help us simplify tricky problems. Think of it as changing the characters in a story to make the plot easier to understand. Here's how it works:
  • Identify a portion of the integral that can be substituted with another variable, called 'u'. This helps transform the integral into a simpler form.
  • Next, compute the derivative of the substituted variable, 'du', to replace 'dx' in the original function.
  • Don't forget to change the limits of integration when you substitute, based on your new variable 'u'.
In our exercise, we used the substitution method by setting \( u = e^x \), which turned the integral into a form easy to integrate: \( \int \frac{u}{1+u^2} du \). This step is crucial for simplifying and solving complex integrals, making them more manageable.
Inverse Trigonometric Functions
Inverse trigonometric functions, like \( \arctan(x) \), play a significant role in calculus, especially in integration. They "undo" what the original trig functions do.In our example:
  • The function \( \arctan(u) \) is the inverse of the tangent function and is used to evaluate the integral \( \int \frac{1}{1+u^2} du \).
  • This is a standard result that comes up frequently, known by many calculus students, where the derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \).
When we use \( \arctan(u) \), we're effectively 'reversing' part of the problem while still keeping it grounded in trigonometric reasoning. In our integral, after substitution had given us \( \frac{u}{1+u^2} \), realising that this was reminiscent of the derivative of an inverse trig function led to our solution.
Limits of Integration
The limits of integration dictate where the integration process starts and stops on the x-axis. When switching variables, as in substitution, these limits must be adjusted too. This maintains the relationship within the new context of the substituted variable.Here's how it affected our example:
  • Originally, the integral's limits were from 0 to 2 for \( x \).
  • Once \( u = e^x \) was introduced, the limits changed to reflect this new variable: \( u \) went from 1 to \( e^2 \).
By reassessing these bounds, we ensure that the area calculation remains consistent, even after transformation.This careful adjustment keeps the integrity of the integral aligning with its initial setup, allowing for accurate area measurement.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Find the position function \(s(t)\) from the given velocity or acceleration function and initial value(s). Assume that units are feet and seconds. $$a(t)=16-t^{2}, v(0)=0, s(0)=30$$

Use the Integral Mean Value Theorem to prove the following fact for a continuous function. For any positive integer \(n\), there exists a set of evaluation points for which the Riemann sum approximation of \(\int_{a}^{b} f(x) d x\) is exact.

Make the indicated substitution for an unspecified function \(f(x)\). $$u=\sin x \text { for } \int_{0}^{\pi / 2}(\cos x) f(\sin x) d x$$

Generalize exercise 51 to \(I=\int_{0}^{d} \frac{f(x)}{f(x)+f(a-x)} d x\) for any positive, continuous function \(f\) and then quickly evaluate \(\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x\)

Derive the formulas \(\int e^{x} d x=e^{x}+c\) and \(\int e^{-x} d x=-e^{-x}+c\)
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.