/*! 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 27 Evaluate the integrals using int... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the integrals using integration by parts where possible. $$\nabla \int\left(x e^{2 x}-4 e^{3 x}\right) d x$$

Short Answer

Expert verified
The short answer for the integral is: \[\int\left(x e^{2 x}-4 e^{3 x}\right) dx = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} - \frac{4}{3} e^{3 x} + C\]

Step by step solution

01

Apply integration by parts to \(x e^{2 x}\)

Let's use the integration by parts formula for the integral of the first term: \[\int u dv = uv - \int v du\] In this case, we assign: \[u = x \quad \Rightarrow \quad du = dx\] \[dv = e^{2 x} dx \quad \Rightarrow \quad v = \frac{1}{2} e^{2 x}\] Now, apply the integration by parts formula: \[\int x e^{2 x} dx = x \cdot \frac{1}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} dx\]
02

Integrate \(\frac{1}{2} e^{2 x} \) in the second term of integration by parts expression

The integral of \(\frac{1}{2} e^{2 x}\) with respect to dx can be found as follows: \[\int \frac{1}{2} e^{2 x} dx = \frac{1}{4} e^{2 x} + C_1\] Now substitute this back into the integration by parts formula: \[x \cdot \frac{1}{2} e^{2 x} - \int \frac{1}{2} e^{2 x} dx = x \cdot \frac{1}{2} e^{2 x} - \left(\frac{1}{4} e^{2 x} + C_1\right)\]
03

Integrate \(-4 e^{3 x}\)

The integral of \(-4 e^{3 x}\) with respect to dx can be found as follows: \[\int -4 e^{3 x} dx = -\frac{4}{3} e^{3 x} + C_2\]
04

Combine both integrals

Now, we will combine both integrals in the expression: \[\int\left(x e^{2 x}-4 e^{3 x}\right) dx = x \cdot \frac{1}{2} e^{2 x} - \left(\frac{1}{4} e^{2 x} + C_1\right) -\frac{4}{3} e^{3 x} + C_2\] We can combine the constants, assigning \(C = C_1 + C_2\), to have the final answer as: \[\int\left(x e^{2 x}-4 e^{3 x}\right) dx = \frac{1}{2} x e^{2 x} - \frac{1}{4} e^{2 x} - \frac{4}{3} e^{3 x} + C\]

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Key Concepts

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

Indefinite Integral
Understanding the indefinite integral is key to solving many calculus problems. An indefinite integral, also known as an antiderivative, is the reverse process of differentiation. If we have a function f(x), the indefinite integral of f(x) with respect to x is a function F(x) such that the derivative of F(x) is f(x). The general form for an indefinite integral is expressed as \[ \int f(x) \, dx = F(x) + C \] where C represents the constant of integration, indicating that there are infinitely many antiderivatives for any given function.

When tackling an exercise like evaluating \( \int (x e^{2 x} - 4 e^{3 x}) dx \), it's important to recall that the solution involves finding an expression for F(x) that, when differentiated, gives us the original integrand. Using the properties of integrals, we can split this integral into two parts for simplicity, but we always have to account for the constant C at the end of our calculations.

The process of integration is more complex when dealing with products of functions, which is where techniques like integration by parts become vital.
Exponential Functions Integration
Exponential functions are frequently encountered in calculus, and knowing how to integrate them is essential. When we have an integrand in the form of an exponential function, such as \( e^{kx} \) where k is a constant, the integral can be derived using a straightforward pattern. The integration of an exponential function can be expressed by the formula: \[ \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C \]

For example, when we integrate \( \frac{1}{2} e^{2x} \) in the exercise, we apply the above rule, keeping in mind that the derivative of \( e^{kx} \) is \( ke^{kx} \) and hence integrating back would mean dividing by that same k. This unwinding step is crucial in solving integration problems involving exponential terms.

Always remember that the power of x in the exponent will affect the constant we divide by in the solution. However, integrating exponential functions also ties closely to integration by parts, especially when they are multiplied with other types of functions, which is the case with the term \( x e^{2x} \) from the exercise.
Integrating Factor Technique
The integrating factor technique is often associated with solving certain types of differential equations, specifically linear first-order differential equations. However, it can be a useful concept to think about when integrating functions, especially as it relates to understanding how multiplying by a cleverly chosen function can simplify an integral.

In our specific context of solving an indefinite integral, the technique mirrors the process of simplification seen in differential equations where we multiply through by an 'integrating factor' to facilitate easier integration. Even though this particular exercise does not directly apply this technique, understanding it can enhance your overall grasp of integration methods.

When dealing with more complex integrals or differential equations, you might find the need to simplify the problem by choosing an appropriate function to multiply throughout your equation, effectively turning the integrand into a form that's easier to manage. While the integrating factor technique is not entirely central to the exercise at hand, it reminds us that, occasionally, multiplying an integrand by a suitable function (or modifying it in some way) can turn a difficult problem into a manageable one.

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Most popular questions from this chapter

For each differential equation, find the particular solution indicated. HINT [See Example 2b.] \(\frac{d y}{d x}=\frac{y^{2}}{x^{2}} ; y=\frac{1}{2}\) when \(x=1\).

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Why is there always an arbitrary constant in the general solution of a differential equation? Why are there not two or more arbitrary constants in a first-order differential equation?

Find a differential equation that is not separable.

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