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Chapter 5: Problem 2

Express the integral in terms of the variable \(u\), but do not evaluate it. (a) \(\int_{-1}^{4}(5-2 x)^{8} d x ; u=5-2 x\) (b) \(\int_{-\pi / 3}^{2 \pi / 3} \frac{\sin x}{\sqrt{2+\cos x}} d x ; u=2+\cos x\) (c) \(\int_{0}^{\pi / 4} \tan ^{2} x \sec ^{2} x d x ; u=\tan x\) (d) \(\int_{0}^{1} x^{3} \sqrt{x^{2}+3} d x ; u=x^{2}+3\)

Short Answer

Expert verified
(a) \( \int_{-3}^{7} \frac{1}{2} u^8 \, du \), (b) \( \int_{3/2}^{5/2} \frac{1}{\sqrt{u}} \, du \), (c) \( \int_{0}^{1} u^2 \, du \), (d) \( \int_{3}^{4} \frac{1}{2} (u-3) \sqrt{u} \, du \).

Step by step solution

01

Express the Integrand in Terms of u for Part (a)

Given u-substitution: let \( u = 5 - 2x \). Differentiate to find \( dx \) in terms of \( du \):\([\frac{du}{dx} = -2] \Rightarrow dx = -\frac{1}{2} du\). Substitute into the integral \( \int (5 - 2x)^8 dx \) becomes \( \int u^8 \cdot \left(-\frac{1}{2}\right) du \).
02

Change the Limit of Integration for Part (a)

Changing variable \( x \) limits to \( u \) limits. For the lower limit when \( x = -1 \), \( u = 5 - 2(-1) = 7 \). For the upper limit when \( x = 4 \), \( u = 5 - 2(4) = -3 \). Adjust the limits accordingly \( \int_{7}^{-3} u^8 \cdot (-\frac{1}{2}) du = \int_{-3}^{7} \frac{1}{2} u^8 du \).
03

Express the Integrand in Terms of u for Part (b)

Given u-substitution: let \( u = 2 + \cos x \). Differentiate: \( \frac{du}{dx} = -\sin x \) which means \( \sin x \, dx = -du \). Substitute \( \int \frac{\sin x}{\sqrt{2+\cos x}} \, dx \) becomes \(-\int \frac{1}{\sqrt{u}} \, du\).
04

Change the Limit of Integration for Part (b)

Changing variable \( x \) limits to \( u \) limits. For the lower limit when \( x = -\frac{\pi}{3} \), \( u = 2 + \cos(-\frac{\pi}{3}) = 2 + \frac{1}{2} = \frac{5}{2} \). For the upper limit when \( x = \frac{2\pi}{3} \), \( u = 2 + \cos(\frac{2\pi}{3}) = 2 - \frac{1}{2} = \frac{3}{2} \). Adjust the limits to get \(-\int_{5/2}^{3/2} \frac{1}{\sqrt{u}} \, du = \int_{3/2}^{5/2} \frac{1}{\sqrt{u}} \, du\).
05

Express the Integrand in Terms of u for Part (c)

Given u-substitution: let \( u = \tan x \). Differentiate: \( \frac{du}{dx} = \sec^2 x \) thus \( \sec^2 x \, dx = du \). Substitute \( \int \tan^2 x \sec^2 x \, dx \) becomes \( \int u^2 \, du \).
06

Change the Limit of Integration for Part (c)

Changing variable \( x \) limits to \( u \) limits. For the lower limit \( x = 0 \), \( u = \tan(0) = 0 \). For the upper limit \( x = \frac{\pi}{4} \), \( u = \tan(\frac{\pi}{4}) = 1 \). Adjust the limits to get \( \int_{0}^{1} u^2 \ du \).
07

Express the Integrand in Terms of u for Part (d)

Given u-substitution: let \( u = x^2 + 3 \). Differentiate: \( \frac{du}{dx} = 2x \) so \( x dx = \frac{1}{2} du \), then \( x^3 dx = x^2 \cdot x dx = \frac{x^2}{2} du \). Express \( x^2 \) in terms of \( u \): \( x^2 = u - 3 \). Substitute into integral \( \int x^3 \sqrt{x^2+3} \ dx \) becomes \( \int \frac{1}{2} (u-3) \sqrt{u} \ du \).
08

Change the Limit of Integration for Part (d)

Changing variable \( x \) limits to \( u \) limits. For lower limit \( x = 0 \), \( u = 0^2 + 3 = 3 \). For upper limit \( x = 1 \), \( u = 1^2 + 3 = 4 \). Adjust the limits to get \( \int_{3}^{4} \frac{1}{2} (u-3) \sqrt{u} \, du \).

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Key Concepts

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

Integral limits
Whenever you're working with definite integrals, the integral limits are very important. They define the start and end of the region over which you're integrating. When performing a substitution, you need to change these limits from the original variable to the substitution variable. This is because the substitution transforms the entire integral into a new variable system. To change the integral limits, evaluate the substitution function at the original boundary values.
  • For example, if the original variable is defined from 0 to 1, and the substitution is given by a function, you plug in these boundary values into the substitution function to derive the new limits.
  • Remember, the direction of limits can also change if the transformed limits are defined from a higher to a lower number.
It's crucial to apply these new limits to the definite integral to ensure correctness.
Trigonometric functions
Trigonometric functions like sine, cosine, and tangent are frequently encountered in calculus and integrals. They often appear in problems that use substitution because they have straightforward derivatives and relationships which make integration easier.
  • The derivative of sin is cos, of cos is -sin, and of tan is sec虏, for example.
  • These derivatives are useful when setting up a substitution involving trigonometric identities.
When solving such integrals, using these relationships can simplify the expressions into basic forms that are easier to integrate once the substitution is complete.
Understanding how to manipulate these functions can make solving complex integrals much more manageable.
Variable substitution
Variable substitution is a technique to simplify complex integrals, often referred to as "u-substitution." It's an essential tool in integration that transforms an integral into a form that is easier to solve. Begin by substituting a part of your original integrand with a new variable, typically called "u." This is often a function inside the integral that complicates integration.
  • For instance, if you see a composition of functions or a trigonometric function that repeats, it might be a good candidate for substitution.
  • Multiply both sides of the substitution by the differential (dx or du), which allows you to replace dx in the integral with something in terms of du.
Finally, remember to adjust your limits of integration to match the new variable. This step is crucial for definite integrals, as it keeps the solution framework consistent and correct.
Definite integrals
Definite integrals are a way to calculate the area under a curve between two boundaries. When applied, they yield a number representing the total accumulated quantity, like area, between these boundaries. A defining characteristic is using upper and lower limits in the integration symbol, indicating the range over which you're integrating. This differs from indefinite integrals, which do not have bounds and yield a family of functions as solutions.
  • When performing definite integrals with a substitution method, ensure you properly transform and apply the limits.
  • The transformed integral may become simpler, but both the form of the integrand and the limits need adjustment.
Using definite integrals correctly identifies a specific value rather than a function plus constant, giving a precise solution to the problem at hand.

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