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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
Question: Evaluate the integral \(\int \frac{4 + e^{-2x}}{e^{3x}} dx\). Answer: \(-\frac{4}{3}e^{-3x} - \frac{1}{5}e^{-5x} + C\)

Step by step solution

01

Rewrite the given expression

Divide each term in the numerator by \(e^{3x}\). We can rewrite the given expression as: $$\int \frac{4+e^{-2 x}}{e^{3 x}} d x = \int \frac{4}{e^{3x}}+ \frac{e^{-2x}}{e^{3x}} d x$$
02

Simplify the terms

Using exponent rules to simplify the terms inside the integral, we get: $$\int \frac{4}{e^{3x}}+ \frac{e^{-2x}}{e^{3x}} d x = \int 4e^{-3x} + e^{-5x} dx$$
03

Separate the integrals and find the antiderivative

We can now rewrite the integral as a sum of two simpler integrals: $$\int 4e^{-3x} + e^{-5x} dx = 4\int e^{-3x} dx + \int e^{-5x} dx$$ Finding the antiderivatives, we get: $$4 \cdot \frac{-1}{3}e^{-3x} + \frac{-1}{5}e^{-5x} = -\frac{4}{3}e^{-3x} - \frac{1}{5}e^{-5x}$$
04

Add the constant of integration

To obtain the general solution, we need to add the constant of integration, \(C\): $$-\frac{4}{3}e^{-3x} - \frac{1}{5}e^{-5x} + C$$ Thus, the integral of the given expression is: $$\int \frac{4+e^{-2 x}}{e^{3 x}} d x = -\frac{4}{3}e^{-3x} - \frac{1}{5}e^{-5x} + C$$

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Most popular questions from this chapter

Apply Simpson's Rule to the following integrals. It is easiest to obtain the Simpson's Rule approximations from the Trapezoid Rule approximations, as in Example \(7 .\) Make \(a\) table similar to Table 7.8 showing the approximations and errors for \(n=4,8,16,\) and \(32 .\) The exact values of the integrals are given for computing the error. \(\int_{0}^{4}\left(3 x^{5}-8 x^{3}\right) d x=1536\)

Evaluate the following integrals or state that they diverge. $$\int_{-2}^{6} \frac{d x}{\sqrt{|x-2|}}$$

Recall that the substitution \(x=a \sec \theta\) implies either \(x \geq a\) (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 \(\tan \theta \leq 0\) ). Graph the function \(f(x)=\frac{1}{x \sqrt{x^{2}-36}}\) on its domain. Then find the area of the region \(R_{1}\) bounded by the curve and the \(x\) -axis on \([-12,-12 / \sqrt{3}]\) and the area of the region \(R_{2}\) bounded by the curve and the \(x\) -axis on \([12 / \sqrt{3}, 12] .\) Be sure your results are consistent with the graph.

Use the reduction formulas in a table of integrals to evaluate the following integrals. $$\int \sec ^{4} 4 t d t$$

Evaluate the following integrals or state that they diverge. $$\int_{-2}^{2} \frac{d p}{\sqrt{4-p^{2}}}$$
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