/*! 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
Answer: The integral of \(\frac{1 - x}{1 - \sqrt{x}} dx\) is \(x + \frac{2x\sqrt{x}}{3} + 2C\), where \(C\) is the constant of integration.

Step by step solution

01

Let u be the square root of x

Let's substitute \(u\) for \(\sqrt{x}\). $$u = \sqrt{x}$$
02

Find du

To find \(du\), differentiate the \(u\) with respect to \(x\) and multiply by \(d x\). $$\frac{d u}{d x} = \frac{1}{2 \sqrt{x}} \implies d u = \frac{1}{2 \sqrt{x}} d x$$ Now, note that this relates \(du\) to \(\sqrt{x}\) and \(dx\), allowing us to substitute into the original integral.
03

Replace x and dx with u

From \(u = \sqrt{x}\), we know that \(x = u^2\). Also, we have \(d u = \frac{1}{2 \sqrt{x}} d x\), which implies \(d x = 2u du\). Substitute these expressions into the integral: $$\int \frac{1 - x}{1 - u} (2u du) = 2 \int \frac{1-u^2}{1-u} u du$$
04

Simplify the integrand

We can simplify the integrand by dividing the numerator and denominator by \((1-u)\): $$2 \int \frac{(1 - u^2) / (1 - u)}{(1 - u) / (1-u)} u du = 2 \int (1+u)u du$$
05

Integrate with respect to u

Now, integrate the simplified expression with respect to \(u\). $$2 \int (1+u)u du = 2\int (u+u^2) du$$ Now integrate each term separately: $$= 2\left[\frac{u^2}{2} + \frac{u^3}{3} + C\right]$$
06

Substitute back x

Since \(u = \sqrt{x}\), we can substitute back to get the final answer. $$= 2\left[\frac{(\sqrt{x})^2}{2} + \frac{(\sqrt{x})^3}{3} + C\right]$$ Simplify the expression: $$= x + \frac{2x\sqrt{x}}{3} + 2C$$ So, the final answer is: $$\int \frac{1-x}{1-\sqrt{x}} d x = x + \frac{2x\sqrt{x}}{3} + 2C$$

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Most popular questions from this chapter

Given a Midpoint Rule approximation \(M(n)\) and a Trapezoid Rule approximation \(T(n)\) for a continuous function on \([a, b]\) with \(n\) subintervals, show that \(T(2 n)=(T(n)+M(n)) / 2\).

The nucleus of an atom is positively charged because it consists of positively charged protons and uncharged neutrons. To bring a free proton toward a nucleus, a repulsive force \(F(r)=k q Q / r^{2}\) must be overcome, where \(q=1.6 \times 10^{-19} \mathrm{C}\) (coulombs) is the charge on the proton, \(k=9 \times 10^{9} \mathrm{N}-\mathrm{m}^{2} / \mathrm{C}^{2}, Q\) is the charge on the nucleus, and \(r\) is the distance between the center of the nucleus and the proton. Find the work required to bring a free proton (assumed to be a point mass) from a large distance \((r \rightarrow \infty)\) to the edge of a nucleus that has a charge \(Q=50 q\) and a radius of \(6 \times 10^{-11} \mathrm{m}\)

For what values of \(p\) does the integral \(\int_{2}^{\infty} \frac{d x}{x \ln ^{p} x}\) exist and what is its value (in terms of \(p\) )?

Apply Simpson's Rule to the following integrals. It is easiest to obtain the Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(7 .\) Make \(a\) table similar to Table 7.8 showing the approximations and errors for \(n=4,8,16,\) and \(32 .\) The exact values of the integrals are given for computing the error. \(\int_{0}^{4}\left(3 x^{5}-8 x^{3}\right) d x=1536\)

Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int p^{2} e^{-3 p} d p$$
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