/*! 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 8 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 8

Evaluate the following integrals or state that they diverge. $$\int_{1}^{\infty} 2^{-x} d x$$

Short Answer

Expert verified
Question: Evaluate the integral \(\int_{1}^{\infty} 2^{-x} d x\). Answer: The value of the integral is \(\frac{1}{2\ln(2)}\).

Step by step solution

01

Determine the antiderivative of the integrand

The given integral is \(\int_{1}^{\infty} 2^{-x} d x\). First, rewrite the integrand as \(2^{-x} = \frac{1}{2^x}\) to make the integration process easier. Now, we can find the antiderivative: $$F(x) = \int \frac{1}{2^x} d x = \int 2^{-x} d x$$ To find the antiderivative, we use the formula for integrating exponential functions: $$\int a^{kx} dx = \frac{a^{kx}}{k \ln(a)} + C$$ Applying this formula to our integrand, we get: $$F(x) = \frac{2^{-x}}{-\ln(2)} + C$$
02

Evaluate the definite integral with the given bounds

Since we have a definite integral, there is no need to worry about the constant of integration, C. Next, we will evaluate the definite integral from \(1\) to \(\infty\): $$\int_{1}^{\infty} 2^{-x} d x = \lim_{b \to \infty} \left[ \frac{2^{-x}}{-\ln(2)} \right]_{1}^{b}$$ Now, substitute the limits of integration and subtract: $$\lim_{b \to \infty} \left( \frac{2^{-b}}{-\ln(2)} - \frac{2^{-1}}{-\ln(2)} \right)$$ As \(b\) approaches infinity, \(2^{-b}\) approaches 0: $$\lim_{b \to \infty} \frac{2^{-b}}{-\ln(2)} = 0$$ Now, substitute this back into our equation: $$0 - \frac{2^{-1}}{-\ln(2)} = \frac{1}{2\ln(2)}$$ Thus, the value of the integral is: $$\int_{1}^{\infty} 2^{-x} d x = \frac{1}{2\ln(2)}$$

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Most popular questions from this chapter

Suppose that a function \(f\) has derivatives of all orders near \(x=0 .\) By the Fundamental Theorem of Calculus, \(f(x)-f(0)=\int_{0}^{x} f^{\prime}(t) d t\) a. Evaluate the integral using integration by parts to show that $$f(x)=f(0)+x f^{\prime}(0)+\int_{0}^{x} f^{\prime \prime}(t)(x-t) d t.$$ b. Show (by observing a pattern or using induction) that integrating by parts \(n\) times gives $$\begin{aligned} f(x)=& f(0)+x f^{\prime}(0)+\frac{1}{2 !} x^{2} f^{\prime \prime}(0)+\cdots+\frac{1}{n !} x^{n} f^{(n)}(0) \\ &+\frac{1}{n !} \int_{0}^{x} f^{(n+1)}(t)(x-t)^{n} d t+\cdots \end{aligned}$$ This expression is called the Taylor series for \(f\) at \(x=0\).

Consider the curve \(y=\ln x\) a. Find the length of the curve from \(x=1\) to \(x=a\) and call it \(L(a) .\) (Hint: The change of variables \(u=\sqrt{x^{2}+1}\) allows evaluation by partial fractions.) b. Graph \(L(a)\) c. As \(a\) increases, \(L(a)\) increases as what power of \(a ?\)

Use symmetry to evaluate the following integrals. a. \(\int_{-\infty}^{\infty} e^{|x|} d x \quad\) b. \(\int_{-\infty}^{\infty} \frac{x^{3}}{1+x^{8}} d x\)

\(\pi < \frac{22}{7}\) One of the earliest approximations to \(\pi\) is \(\frac{22}{7} .\) Verify that \(0 < \int_{0}^{1} \frac{x^{4}(1-x)^{4}}{1+x^{2}} d x=\frac{22}{7}-\pi .\) Why can you conclude that \(\pi < \frac{22}{7} ?\)

Prove the following orthogonality relations (which are used to generate Fourier series). Assume \(m\) and \(n\) are integers with \(m \neq n\) a. \(\int_{0}^{\pi} \sin m x \sin n x d x=0\) b. \(\int_{0}^{\pi} \cos m x \cos n x d x=0\) c. \(\int_{0}^{\pi} \sin m x \cos n x d x=0\)
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