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Chapter 5: Problem 50

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{1 / 4} \frac{x}{\sqrt{1-16 x^{2}}} d x$$

Short Answer

Expert verified
Answer: The value of the definite integral is $$\frac{1}{2}.$$

Step by step solution

01

Choose the substitution

Let's make a substitution for the expression under the square root. A common substitution choice in this scenario is to use trigonometric variables. Let's choose: $$u = 4x\Rightarrow x=\frac{u}{4}$$ Now differentiate both sides to find \(dx\) in terms of \(du\). $$\frac{d x}{d u}=\frac{1}{4} \Rightarrow d x=\frac{1}{4} d u$$
02

Change the limits of integration

Now, we need to change the given limits of integration from \(x\) to \(u\). Using our substitution \(x=\frac{u}{4}\), we will find the corresponding \(u\) values for \(x=0\) and \(x=\frac{1}{4}\): For \(x=0\): $$u = 4\cdot0 = 0$$ For \(x=\frac{1}{4}\): $$u = 4\cdot\frac{1}{4} = 1$$ So, the new limits of integration are \(0\) and \(1\) for \(u\).
03

Rewrite the integral

Now, we need to rewrite the given integral in terms of \(u\). We have: $$\int_{0}^{1 / 4} \frac{x}{\sqrt{1-16 x^{2}}} d x = \int_{0}^{1} \frac{\frac{u}{4}}{\sqrt{1-16(\frac{u}{4})^2}} \cdot \frac{1}{4} du$$ Simplifying the expression gives: $$\int_{0}^{1} \frac{u}{\sqrt{1 - u^2}} du$$
04

Solve the integral

Now, we can evaluate the integral: $$\int_{0}^{1} \frac{u}{\sqrt{1 - u^2}} du$$ This integral is a standard trigonometric substitution integral, and its antiderivative is: $$-\frac{1}{2}\sqrt{1-u^2}+C$$ Now we evaluate the definite integral from \(0\) to \(1\): $$\left[-\frac{1}{2}\sqrt{1-u^2}\right]_{0}^{1} = -\frac{1}{2}\sqrt{1-1^2} + \frac{1}{2}\sqrt{1-0^2}= -0 + \frac{1}{2} = \frac{1}{2}$$ Therefore, the value of the definite integral is: $$\int_{0}^{1 / 4} \frac{x}{\sqrt{1-16 x^{2}}} d x = \frac{1}{2}$$

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Most popular questions from this chapter

Additional integrals Use a change of variables to evaluate the following integrals. $$\int \sec 4 w \tan 4 w d w$$

Use the definition, of the definite integral to justify the property \(\int_{a}^{b} c f(x) d x=c \int_{a}^{b} f(x) d x,\) where \(f\) is continuous and \(c\) is a real number.

Evaluate the following definite integrals using the Fundamental Theorem of Calculus. $$\int_{0}^{\sqrt{3}} \frac{3 d x}{9+x^{2}}$$

Consider the function \(f\) and the points \(a, b,\) and \(c\) a. Find the area function \(A(x)=\int_{a}^{x} f(t) d t\) using the Fundamental Theorem. b. Graph \(f\) and \(A\) c. Evaluate \(A(b)\) and \(A(c)\) and interpret the results using the graphs of part \((b)\) $$f(x)=\cos \pi x ; a=0, b=\frac{1}{2}, c=1$$

\(\sin ^{2} a x\) and \(\cos ^{2} a x\) integrals Use the Substitution Rule to prove that $$\begin{array}{l}\int \sin ^{2} a x d x=\frac{x}{2}-\frac{\sin (2 a x)}{4 a}+C \text { and } \\\\\int \cos ^{2} a x d x=\frac{x}{2}+\frac{\sin (2 a x)}{4 a}+C\end{array}$$
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