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Chapter 7: Problem 12

Evaluate the following integrals. $$\int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x$$

Short Answer

Expert verified
Based on the step-by-step solution, calculate the values of $$F(1)$$ and $$F(\frac{1}{2})$$ and write the final expression for the definite integral $$\int_{1/2}^1 \frac{\sqrt{1-x^2}}{x^2}dx$$. Answer: $$F(1) = \frac{1}{2}$$ $$F\left(\frac{1}{2}\right) = -\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\ln\left(\frac{1}{4}\right) + \int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x\right)$$ So the definite integral is: $$\int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x = \frac{1}{2} +\frac{1}{2}\left(\frac{\sqrt{3}}{2}\ln\left(\frac{1}{4}\right)-\int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x\right)$$

Step by step solution

01

Find the antiderivative of the given function

First, let's find the antiderivative of the function $$\frac{\sqrt{1-x^{2}}}{x^{2}}.$$ To do this, we will use a substitution method. Let's use the substitution $$u = 1-x^2$$ which implies that $$du = -2xdx$$. So we also have $$-\frac{1}{2}du = xdx$$ Now, we have: $$\int \frac{\sqrt{1-x^{2}}}{x^{2}} d x=\int \frac{\sqrt{u}}{x^{2}}\left(-\frac{1}{2}\right) d u$$
02

Evaluate the new integral

Next, we need to replace $$x^2$$ in the denominator with an expression in terms of $$u$$. From our substitution, we can see that $$x^2 = 1-u$$. Now we have: $$\int \frac{\sqrt{1-x^{2}}}{x^{2}} d x= -\frac{1}{2} \int \frac{\sqrt{u}}{1-u} d u$$
03

Apply integration by parts

Now we will apply integration by parts to evaluate the integral. Let $$dv=\frac{1}{1-u} du$$, then $$v=-\ln|1-u|$$ and let $$u=\sqrt{u}$$, then $$du=\frac{1}{2\sqrt{u}} du$$. Applying integration by parts formula, we get: $$-\frac{1}{2}\left(-\sqrt{u}\ln|1-u|-\int\frac{\sqrt{u}du}{1-u}\right)$$
04

Solve the remaining integral and evaluate final expression

Now let's go back to the original variable, $$x$$. Remember that $$u = 1-x^2$$, so $$F(x)=-\frac{1}{2}\left(-\sqrt{1-x^{2}}\ln|{x^{2}}|+\int\frac{\sqrt{1-x^{2}}dx}{x^{2}}\right)$$. Now we need to evaluate the definite integral: $$\int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x = F(1)-F\left(\frac{1}{2}\right)$$
05

Calculate the values

Plug in $$x = 1$$ and $$x = \frac{1} {2}$$ into the antiderivative to find their values: $$F(1)=-\frac{1}{2}(0-1) =\frac{1}{2}$$ $$F\left(\frac{1}{2}\right)=-\frac{1}{2}\left(-\frac{\sqrt{3}}{2}\ln\left(\frac{1}{4}\right) + \int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x\right)$$ Now, plugging these values back into the definite integral expression: $$\int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x =\frac{1}{2} +\frac{1}{2}\left(\frac{\sqrt{3}}{2}\ln\left(\frac{1}{4}\right)-\int_{1 / 2}^{1} \frac{\sqrt{1-x^{2}}}{x^{2}} d x\right)$$ Note that we can't evaluate this integral exactly, and we would typically use a numerical method to approximate the value.

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Most popular questions from this chapter

Suppose \(f\) is positive and its first two derivatives are continuous on \([a, b] .\) If \(f^{\prime \prime}\) is positive on \([a, b],\) then is a Trapezoid Rule estimate of \(\int_{a}^{b} f(x) d x\) an underestimate or overestimate of the integral? Justify your answer using Theorem 2 and an illustration.

Shortcut for the Trapezoid Rule Prove that if you have \(M(n)\) and \(T(n)\) (a Midpoint Rule approximation and a Trapezoid Rule approximation with \(n\) subintervals), then \(T(2 n)=(T(n)+M(n)) / 2\).

\(A\) powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function \(f(t),\) the Laplace transform is a new function \(F(s)\) defined by $$ F(s)=\int_{0}^{\infty} e^{-s t} f(t) d t $$ where we assume that s is a positive real number. For example, to find the Laplace transform of \(f(t)=e^{-t},\) the following improper integral is evaluated: $$ F(s)=\int_{0}^{\infty} e^{-s t} e^{-t} d t=\int_{0}^{\infty} e^{-(s+1) t} d t=\frac{1}{s+1} $$ Verify the following Laplace transforms, where a is a real number. $$f(t)=e^{a t} \longrightarrow F(s)=\frac{1}{s-a}$$

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\(A\) total charge of \(Q\) is distributed uniformly on a line segment of length \(2 L\) along the \(y\) -axis (see figure). The \(x\) -component of the electric field at a point \((a, 0)\) on the \(x\) -axis is given by $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ where \(k\) is a physical constant and \(a>0\) a. Confirm that \(E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}\) b. Letting \(\rho=Q / 2 L\) be the charge density on the line segment, show that if \(L \rightarrow \infty,\) then \(E_{x}(a)=2 k \rho / a\)
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