/*! 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 29 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 29

Evaluate the following integrals. $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x$$

Short Answer

Expert verified
Question: Calculate the integral: $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x$$ Answer: $$-\sqrt{16-x^2} + C$$

Step by step solution

01

What substitution to use?

We are given the integral: $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x$$ To simplify it, we will use the substitution method. We'll use: $$u = 16 - x^2\Rightarrow d u = -2 x dx$$
02

Substituting the variables

Now we substitute \(u\) for \(16-x^2\) and \(du\) for \(-2x\,dx\) to solve the integral in terms of \(u\): $$\int \frac{x^2}{\sqrt{16 - x^2}} d x = -\frac{1}{2}\int \frac{1}{\sqrt{u}} du$$
03

Integrate

Now, we can integrate the expression with respect to \(u\): $$-\frac{1}{2}\int \frac{1}{\sqrt{u}} d u= -\frac{1}{2}\int u^{-\frac{1}{2}} d u$$ Applying the power rule for integration: $$-\frac{1}{2}\int u^{-\frac{1}{2}} d u = -\frac{1}{2}\cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C$$ After simplifying the expression, we get: $$-\frac{1}{2}\cdot \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C= -u^{\frac{1}{2}} + C$$
04

Replacing the variable

Now we can replace \(u\) back with its original definition \(u = 16-x^2\): $$ -u^{\frac{1}{2}} + C = -\sqrt{16-x^2} + C $$ And this is the final result of our integral: $$\int \frac{x^{2}}{\sqrt{16-x^{2}}} d x = -\sqrt{16-x^2} + C$$

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Most popular questions from this chapter

Challenge Show that with the change of variables \(u=\sqrt{\tan x}\) the integral \(\int \sqrt{\tan x} d x\) can be converted to an integral amenabl to partial fractions. Evaluate \(\int_{0}^{\pi / 4} \sqrt{\tan x} d x\)

Recall that the substitution \(x=a \sec \theta\) implies that \(x \geq a\) (in which case \(0 \leq \theta<\pi / 2\) and \(\tan \theta \geq 0\) ) or \(x \leq-a\) (in which case \(\pi / 2<\theta \leq \pi\) and \(\tan \theta \leq 0\) ). $$ \begin{array}{l} \text { Show that } \int \frac{d x}{x \sqrt{x^{2}-1}}= \\ \qquad\left\\{\begin{array}{ll} \sec ^{-1} x+C=\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x>1 \\ -\sec ^{-1} x+C=-\tan ^{-1} \sqrt{x^{2}-1}+C & \text { if } x<-1 \end{array}\right. \end{array} $$

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 $$\sin x=2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) \text { and } \cos x=2 \cos ^{2}\left(\frac{x}{2}\right)-1$$

On the interval \([0,2],\) the graphs of \(f(x)=x^{2} / 3\) and \(g(x)=x^{2}\left(9-x^{2}\right)^{-1 / 2}\) have similar shapes. a. Find the area of the region bounded by the graph of \(f\) and the \(x\) -axis on the interval [0,2] b. Find the area of the region bounded by the graph of \(g\) and the \(x\) -axis on the interval [0,2] c. Which region has the greater area?

The work required to launch an object from the surface of Earth to outer space is given by \(W=\int_{R}^{\infty} F(x) d x,\) where \(R=6370 \mathrm{km}\) is the approximate radius of Earth, \(F(x)=G M m / x^{2}\) is the gravitational force between Earth and the object, \(G\) is the gravitational constant, \(M\) is the mass of Earth, \(m\) is the mass of the object, and \(G M=4 \times 10^{14} \mathrm{m}^{3} / \mathrm{s}^{2}.\) a. Find the work required to launch an object in terms of \(m.\) b. What escape velocity \(v_{e}\) is required to give the object a kinetic energy \(\frac{1}{2} m v_{e}^{2}\) equal to \(W ?\) c. The French scientist Laplace anticipated the existence of black holes in the 18th century with the following argument: If a body has an escape velocity that equals or exceeds the speed of light, \(c=300,000 \mathrm{km} / \mathrm{s},\) then light cannot escape the body and it cannot be seen. Show that such a body has a radius \(R \leq 2 G M / c^{2} .\) For Earth to be a black hole, what would its radius need to be?
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