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Evaluate the integrals using integration by parts where possible. $$\nabla \frac{2 x+1}{e^{3 x}} d x$$

Short Answer

Expert verified
The integral \(\int{\frac{2x + 1}{e^{3x}} dx}\) can be evaluated using integration by parts, with \(u = 2x + 1\) and \(dv = e^{-3x} dx\). The result is \(-\frac{2x}{3} e^{-3x} - \frac{1}{9} e^{-3x} + C\).

Step by step solution

01

Choose the function to differentiate and integrate

For our integrand \(\frac{2x + 1}{e^{3x}}\), we'll differentiate the polynomial part, \(u = 2x + 1\), and integrate the exponential part, \(dv = e^{-3x} dx\).
02

Differentiate u and integrate dv

Differentiate \(u = 2x + 1\) with respect to x, and integrate \(dv = e^{-3x} dx\) with respect to x: \(du = (2) dx\) \(v = -\frac{1}{3} e^{-3x}\)
03

Apply the integration by parts formula

Now we need to apply the integration by parts formula, which states that: \(\int{u*dv} = u*v - \int{v*du}\) Substitute our expressions for u, du, v, and dv obtained from Step 2: \(\int{\frac{2x + 1}{e^{3x}} dx} = (2x + 1)(-\frac{1}{3} e^{-3x}) - \int{-\frac{1}{3} e^{-3x}(2) dx}\)
04

Simplify the expression

Now we will simplify the expression and proceed to evaluate the remaining integral: \(-\frac{2x}{3} e^{-3x} - \frac{1}{3} e^{-3x} + \frac{2}{3} \int{e^{-3x} dx}\)
05

Integrate the remaining part and add the constant of integration

Now we just need to integrate the last remaining exponential part and add the constant of integration: \(-\frac{2x}{3} e^{-3x} - \frac{1}{3} e^{-3x} + \frac{2}{9} e^{-3x} + C\)
06

Write the final answer

Combine the terms to get the final answer: \(\int{\frac{2x + 1}{e^{3x}} dx} = -\frac{2x}{3} e^{-3x} - \frac{1}{9} e^{-3x} + C\)

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Key Concepts

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

Exponential Functions
Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a constant base and \(x\) is the variable exponent. However, in calculus, the most common base is the natural number \(e\), approximately equal to 2.718. These functions are crucial in modeling growth and decay processes.
When approaching integration or differentiation involving exponential functions, knowing their unique properties can simplify the process.
  • The derivative of \(e^{kx}\) is \(ke^{kx}\), where \(k\) is a constant.
  • The integral of \(e^{kx}\) with respect to \(x\) is \(\frac{1}{k}e^{kx} + C\).
In the integration by parts example, the exponential function used is \(e^{-3x}\). Noticing its behavior helps in determining how it interacts with other parts of the integrand. Specifically, it decays to zero as \(x\) increases, which often makes it 'nicer' to integrate and differentiate.
Polynomial Functions
Polynomial functions are expressions involving a sum of powers of \(x\) with coefficients. They are one of the most fundamental types of functions in mathematics and can be written in the form \(a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0\).
When integrating polynomial functions, each term is handled by increasing the power by one and dividing by the new power:
  • The integral of \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), as long as \(neq -1\).
In the specific problem given, the polynomial function \(2x + 1\) is being integrated. The rule of thumb in integration by parts is to differentiate polynomial parts since they simplify quickly, breaking down term by term. Here, differentiating \(2x + 1\) results in the constant \(2\), taking us further towards solving the integral.
Definite and Indefinite Integrals
Integrals can be categorized as definite or indefinite. In indefinite integrals, we seek a general form of the antiderivative without any bounds of integration. The notation for an indefinite integral is \(\int f(x) \, dx = F(x) + C\), where \(F(x)\) is an antiderivative of the function \(f(x)\) and \(C\) is the constant of integration.
In definite integrals, however, specific boundaries \([a, b]\) are set, and we're interested in calculating the net area between the curve and the x-axis from \(a\) to \(b\):
  • \(\int_{a}^{b} f(x) \, dx = F(b) - F(a)\)
In the exercise presented here, an indefinite integral is evaluated because no limits are provided. The constant \(C\) is added to represent the family of curves that differ only vertically. This constant ensures that all potential initial conditions for a differential equation can be accounted for in the solution.

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Most popular questions from this chapter

Electric Circuits The flow of current \(i(t)\) in an electric circuit without capacitance satisfies the linear differential equation $$ L \frac{d i}{d t}+R i=V(t) $$ where \(L\) and \(R\) are constants (the inductance and resistance respectively) and \(V(t)\) is the applied voltage. (See figure.) If the voltage is supplied by a 10 -volt battery and the switch is turned on at time \(t=1\), then the voltage \(V\) is a step function that jumps from 0 to 10 at \(t=1: V(t)=5\left[1+\frac{|t-1|}{t-1}\right]\). Find the current as a function of time for \(L=R=1 .\) Use a grapher to plot the resulting current as a function of time. (Assume there is no current flowing at time \(t=0 .\) [Use the following integral formula: \(\int\left[1+\frac{|t-1|}{t-1}\right] e^{t} d t=\) \(\left.\left[1+\frac{|t-1|}{t-1}\right]\left(e^{t}-e\right)+C .\right]\)

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