/*! 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 26 Evaluate the following integrals... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 26

Evaluate the following integrals. $$\int \frac{4+e^{-2 x}}{e^{3 x}} d x$$

Short Answer

Expert verified
Based on the step by step solution, the integral $$\int \frac{4+e^{-2 x}}{e^{3 x}} d x$$ evaluates to $$4e^x - e^{-x} + C$$

Step by step solution

01

Simplify the integrand

We are given the integral: $$\int \frac{4+e^{-2 x}}{e^{3 x}} d x$$ We can rewrite the numerator by factoring out \(e^{-2 x}\): $$\int \frac{e^{-2 x}(4 e^{2 x} + 1)}{e^{3 x}} d x$$ Now we can cancel out the common term \(e^{-2 x}\) from numerator and denominator: $$\int \frac{4 e^{2x} + 1}{e^x} d x$$
02

Separate the integrand

We can separate the integrand into two parts: $$\int \frac{4 e^{2x}}{e^x} d x + \int \frac{1}{e^x} d x$$ $$=\int 4e^x d x + \int e^{-x} d x$$
03

Integrate each part

Now we can integrate each part separately: 1. \(\int 4e^x d x\): To integrate \(4e^x d x\), we simply find the antiderivative of \(4e^x\): $$4 \int e^x d x = 4 e^x$$ 2. \(\int e^{-x} d x\): To integrate \(e^{-x} d x\), we find the antiderivative of \(e^{-x}\): $$\int e^{-x} d x = - e^{-x}$$
04

Combine the results and add the constant of integration

Now we combine the results from step 3 and add the constant of integration, \(C\): $$4e^x - e^{-x} + C$$ Thus, the integral evaluates to: $$\int \frac{4+e^{-2 x}}{e^{3 x}} d x = 4e^x - e^{-x} + C$$

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Most popular questions from this chapter

Imagine that today you deposit \(\$ B\) in a savings account that earns interest at a rate of \(p \%\) per year compounded continuously. The goal is to draw an income of \(\$ I\) per year from the account forever. The amount of money that must be deposited is \(B=I \int_{0}^{\infty} e^{-n t} d t,\) where \(r=p / 100 .\) Suppose you find an account that earns \(12 \%\) interest annully and you wish to have an income from the account of \(\$ 5000\) per year. How much must you deposit today?

Evaluate the following integrals. Consider completing the square. $$\int_{2+\sqrt{2}}^{4} \frac{d x}{\sqrt{(x-1)(x-3)}}$$

Use the following three identities to evaluate the given integrals. $$\begin{aligned}&\sin m x \sin n x=\frac{1}{2}[\cos ((m-n) x)-\cos ((m+n) x)]\\\&\sin m x \cos n x=\frac{1}{2}[\sin ((m-n) x)+\sin ((m+n) x)]\\\&\cos m x \cos n x=\frac{1}{2}[\cos ((m-n) x)+\cos ((m+n) x)]\end{aligned}$$ $$\int \sin 5 x \sin 7 x d x$$

Use the Trapezoid Rule (Section 7 ) to approximate \(\int_{0}^{R} e^{-x^{2}} d x\) with \(R=2,4,\) and 8. For each value of \(R\), take \(n=4,8,16,\) and \(32,\) and compare approximations with successive values of \(n .\) Use these approximations to approximate \(I=\int_{0}^{\infty} e^{-x^{2}} d x.\)

Let \(R\) be the region between the curves \(y=e^{-c x}\) and \(y=-e^{-c x}\) on the interval \([a, \infty),\) where \(a \geq 0\) and \(c \geq 0 .\) The center of mass of \(R\) is located at \((\bar{x}, 0)\) where \(\bar{x}=\frac{\int_{a}^{\infty} x e^{-c x} d x}{\int_{a}^{\infty} e^{-c x} d x} .\) (The profile of the Eiffel Tower is modeled by the two exponential curves.) a. For \(a=0\) and \(c=2,\) sketch the curves that define \(R\) and find the center of mass of \(R\). Indicate the location of the center of mass. b. With \(a=0\) and \(c=2,\) find equations of the lines tangent to the curves at the points corresponding to \(x=0.\) c. Show that the tangent lines intersect at the center of mass. d. Show that this same property holds for any \(a \geq 0\) and any \(c>0 ;\) that is, the tangent lines to the curves \(y=\pm e^{-c x}\) at \(x=a\) intersect at the center of mass of \(R\) (Source: P. Weidman and I. Pinelis, Comptes Rendu, Mechanique \(332(2004): 571-584 .)\)
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