/*! 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 38 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 38

Evaluate the following integrals. $$\int \frac{1-x}{1-\sqrt{x}} d x$$

Short Answer

Expert verified
#tag_title# Solution #tag_content# The solution to the given integral is: $$2((1 - \sqrt{x}) - \frac{1}{3}(1 - \sqrt{x})^3) + C$$

Step by step solution

01

Perform substitution

Let \(u = 1 - \sqrt{x}\). Then, we will differentiate both sides with respect to x: $$\frac{d u}{d x} = -\frac{1}{2\sqrt{x}}$$ Now, we will solve for \(\frac{dx}{du}\): $$\frac{d x}{d u} = -2\sqrt{x}$$ We can express the x-term in the numerator with u by solving for x in terms of u: $$x = (1 - u)^2$$ Now, we rewrite the integral in terms of u:
02

Rewrite in terms of u

Substituting the expressions derived above, the integral becomes: $$\int \frac{1 - (1 - u)^2}{u} (-2\sqrt{(1 - u)^2}) d u$$ Simplify the integrand:
03

Simplify the integrand

The integrand simplifies to: $$\int 2 (1 - u^2) d u$$ Now, we will integrate with respect to u:
04

Integrate with respect to u

Integrating the expression, we get: $$2 \int (1 - u^2) d u = 2(u - \frac{1}{3} u^3) + C$$ where C is the constant of integration.
05

Substitute back x

Substitute the original expression for u back, that is, \(u = 1 - \sqrt{x}\): $$2((1 - \sqrt{x}) - \frac{1}{3}(1 - \sqrt{x})^3) + C$$ Finally, to get the solution of the given integral, simplify the above expression.

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Most popular questions from this chapter

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{d x}{e^{x}+e^{2 x}}$$

An 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 \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ $$\text { Evaluate } \int \frac{d x}{1+\sin x}$$

Show that \(L=\lim _{n \rightarrow \infty}\left(\frac{1}{n} \ln n !-\ln n\right)=-1\) in the following steps. a. Note that \(n !=n(n-1)(n-2) \cdots 1\) and use \(\ln (a b)=\ln a+\ln b\) to show that $$ \begin{aligned} L &=\lim _{n \rightarrow \infty}\left[\left(\frac{1}{n} \sum_{k=1}^{n} \ln k\right)-\ln n\right] \\ &=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right) \end{aligned} $$ b. Identify the limit of this sum as a Riemann sum for \(\int_{0}^{1} \ln x d x\) Integrate this improper integral by parts and reach the desired conclusion.

The following integrals require a preliminary step such as long division or a change of variables before using partial fractions. Evaluate these integrals. $$\int \frac{2 x^{3}+x^{2}-6 x+7}{x^{2}+x-6} d x$$

An 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 \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. $$A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}$$ Verify relation \(A\) by differentiating \(x=2 \tan ^{-1} u\). Verify relations \(B\) and \(C\) using a right-triangle diagram and the double-angle formulas $$\sin x=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1$$
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