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Chapter 28: Problem 17

Integrate each of the given functions. $$\int_{1}^{3} 3 e^{2 x}\left(e^{-2 x}-1\right) d x$$

Short Answer

Expert verified
The result of the integral is: \(6 - \frac{3}{2}(e^6 - e^2)\).

Step by step solution

01

Distribute the Multiplier

Let's first distribute the constant 3 inside the integral by applying it to both terms within the parentheses. \[\int_{1}^{3} 3e^{2x}(e^{-2x} - 1) dx = \int_{1}^{3} 3e^{2x} \cdot e^{-2x} dx - \int_{1}^{3} 3e^{2x} dx\]
02

Simplify Exponents

Each term in the integral can be further simplified. The first term involves the product of exponentials:\[3e^{2x} \cdot e^{-2x} = 3e^{0} = 3\]So, the first integral becomes a constant function.The entire integral becomes:\[\int_{1}^{3} 3 \, dx - \int_{1}^{3} 3e^{2x} \, dx\]
03

Integrate the Constant Function

Integrate the constant function 3 over the interval from 1 to 3. This simplifies to:\[3 \int_{1}^{3} dx = 3[x]_{1}^{3} = 3(3-1) = 6\]
04

Integrate the Exponential Function

Integrate the function \(3e^{2x}\) from 1 to 3. We need to account for the coefficient from the chain rule:\[\int 3e^{2x} \, dx = \frac{3}{2}e^{2x} + C\]Evaluate this from 1 to 3:\[\frac{3}{2} [e^{2x}]_{1}^{3} = \frac{3}{2}(e^6 - e^2)\]
05

Subtract the Integrals

Now, subtract the result from Step 4 from the result of Step 3:\[6 - \frac{3}{2}(e^6 - e^2)\]Simplify this to find the final answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distributing Constants
Distributing constants is a fundamental step when integrating, especially when dealing with products. In this exercise, we encounter a multiplier outside the parentheses, which needs to be distributed properly inside. Here, we start with the expression \(3e^{2x}(e^{-2x} - 1)\). By distributing the 3 (constant) to each term inside the parentheses, we simplify our integration process.
  • Distribute the 3 to \(e^{2x} imes e^{-2x}\)
  • Then distribute the 3 to \(-1\)
This separation ensures we handle each term individually, making future steps like integration much simpler.
Simplifying Exponents
Exponential expressions often appear complex but can be simplified using basic rules of exponents. Consider \(3e^{2x} imes e^{-2x}\). According to exponent rules:
  • The product of powers with the same base is the base raised to the sum of the exponents. So, \(e^{2x} imes e^{-2x} = e^{0} = 1\).
  • Thus, \(3e^{2x} \cdot e^{-2x}\) simplifies to 3.
This simplification turns the integral part just consisting \(3\), which is a constant and much easier to integrate.
Integrating Constant Functions
When integrating constant functions, it is relatively straightforward. A constant function, like 3, over an interval such as from 1 to 3, requires simple arithmetic. We use the formula for integrating constants: \[ \int_{a}^{b} c \, dx = c[x]_{a}^{b} \]In this case, \(c = 3\), so integrating from 1 to 3 becomes:
  • Calculate \(3[x]_{1}^{3} = 3(3 - 1) = 6\)
  • This is the result for the first simplified integral term.
Constant integrations often serve as quick calculations, aiding in solving composite integrals efficiently.
Integrating Exponential Functions
Exponential functions, while complex, become manageable with proper techniques. Here, we integrate \(3e^{2x}\). First, we use the chain rule from calculus:
  • When integrating \(3e^{2x}\), note the inner derivative \(d/dx(2x) = 2\).
  • We adjust for this by introducing a factor, leading to \(\frac{3}{2}e^{2x}\).
This adjustment aligns the integral with the antiderivative. Finally, evaluate between the limits, 1 and 3: \[\frac{3}{2}[e^{2x}]_{1}^{3} = \frac{3}{2}(e^6 - e^2)\]This covers the exponential component accurately within the original problem's limits.
Definite Integrals
Definite integrals provide a numerical value representing the area under a curve, between specific bounds. By combining the results of our calculations, we subtract the exponential integral from the constant's result:
  • The first term gave us 6.
  • The second term resulted in \(\frac{3}{2}(e^6 - e^2)\).
Our final step involves subtraction: \[6 - \frac{3}{2}(e^6 - e^2)\]This simplification and arithmetic yield the problem's solution. Evaluating definite integrals solidifies understanding by providing complete analysis for the given integral.

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Most popular questions from this chapter

Solve the given problems by integration. In determining the temperature that is absolute zero (0 \(\mathrm{K}\), or about \(\left.-273^{\circ} \mathrm{C}\right),\) the equation \(\ln T=-\int \frac{d r}{r-1}\) is used. Here, \(T\) is the thermodynamic temperature and \(r\) is the ratio between certain specific vapor pressures. If \(T=273.16 \mathrm{K}\) for \(r=1.3361,\) find \(T\) as a function of \(r\) (if \(r>1\) for all \(T\) ).

Certain algebraic integrals can be transformed into integrable form with the appropriate algebraic substitution. For an expression of the form \((a x+b)^{p / q}, a\) substitution of the form \(u=(a x+b)^{1 / q}\) may put it into an integrable form. In Exercises \(39-42,\) use this type of substitution for the given integrals. $$\int x \sqrt{x+1} d x$$

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Solve the given problems by integration. Under ideal conditions, the natural law of population growth is that population increases at a rate proportional to the population \(P\) present at any time \(t .\) This leads to the equation \(t=\frac{1}{k} \int \frac{d P}{P}\). Assuming ideal conditions for the United States, if \(P=249\) million in \(1990(t=0)\) and \(P=275\) million in \(2000(t=10 \text { years })\) find the population that is projected in \(2020(t=30\) years).

$$\text {Solve the given problems by integration.}$$ The displacement \(y\) (in cm) of a weight on a spring is given by \(y=4 e^{-t} \cos t(t \geq 0) .\) Find the average value of the displacement for the interval \(0 \leq t \leq 2 \pi\) s.
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