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Use the reduction formulas in to evaluate the following integrals. $$\int x^{2} e^{3 x} d x$$

Short Answer

Expert verified
Question: Determine the integral of the function \(x^2 e^{3x}\). Answer: The integral of \(x^2 e^{3x}\) is \(\frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \frac{4}{81} e^{3x} + C\), where C is the constant of integration.

Step by step solution

01

Apply integration by parts

To apply integration by parts, we will need two functions, u and v. Here, let's choose: $$u = x^{2}$$ $$dv = e^{3x}dx$$ Next, we need to find the derivatives and integrals of our chosen functions u and v: $$du = 2x dx$$ $$v = \frac{1}{3}e^{3x}$$ Now, apply the integration by parts formula, $$\int u dv = u v - \int v du$$: $$\int x^{2} e^{3 x} d x = \frac{1}{3}x^{2} e^{3x} - \frac{2}{3} \int x e^{3x} dx$$ Now we still have to integrate the new integral: $$\frac{2}{3}\int x e^{3x} dx$$. Since it's still a product between a polynomial and an exponential function, we'll need to apply integration by parts again.
02

Apply integration by parts again

For the new integral, let's choose: $$u = x$$ $$dv = e^{3x}dx$$ Now, find the derivatives and integrals of our chosen functions u and v: $$du = dx$$ $$v = \frac{1}{3}e^{3x}$$ Then, apply the integration by parts formula again: $$\frac{2}{3} \int x e^{3x} dx = \frac{2}{3} \left( \frac{1}{3} x e^{3x} - \frac{1}{3}\int e^{3x} dx \right)$$ Now we need to integrate the remaining integral: $$\frac{2}{9} \int e^{3x} dx$$.
03

Integrate the remaining integral

Integrating the remaining integral is simple: $$\frac{2}{9} \int e^{3x} dx = \frac{2}{9} \cdot \frac{1}{3} e^{3x} + C = \frac{2}{27} e^{3x} + C$$
04

Combine all the expressions

Now, combining the results from steps 1, 2, and 3, we get the final solution: $$\int x^{2} e^{3 x} d x = \frac{1}{3}x^{2} e^{3x} - \frac{2}{3} \left( \frac{1}{3} x e^{3x} - \frac{1}{3}\left(\frac{2}{27}e^{3x}+C\right) \right)$$ $$\int x^{2} e^{3 x} d x = \frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \frac{4}{81} e^{3x} + C$$ The solution is: $$\int x^{2} e^{3 x} d x = \frac{1}{3} x^{2} e^{3x} - \frac{2}{9} x e^{3x} + \frac{4}{81} e^{3x} + C$$

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Key Concepts

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

Reduction Formulas
Reduction formulas are a set of equations in integral calculus that simplify the process of integrating complex functions. They essentially 'reduce' the power of the function term by term until an integral is attained that is easy to solve directly. The method often involves applying integration techniques recursively.

For example, when integrating a function of the form \( x^n e^{kx} \), where \( n \) is an integer and \( k \) is a constant, reduction formulas become extremely useful. We break down the problem by integrating by parts and expressing the more complex integral in terms of simpler ones, which ultimately allows us to find the solution step by step. In the solution provided, the reduction formula approach helped to break down the original integral \(\int x^{2} e^{3 x} d x\) to a point where the integration could be performed without much difficulty.
Exponential Functions
An exponential function is a mathematical expression in the form \(e^{kx}\), where \(e\) is the base of natural logarithms, \(k\) is a constant, and \(x\) is the variable. These functions are known for their unique property where the rate of growth is directly proportional to the value of the function.

In the context of our integral, \(e^{3x}\) is an exponential function where the rate of growth increases threefold for each unit increase in \(x\). The key to integrating exponential functions lies in recognizing patterns and understanding that the integral of \(e^{kx}\) is \(\frac{1}{k}e^{kx}\) plus the constant of integration, \(C\). This principle was applied within the step by step solution multiple times as part of the integration by parts process.
Indefinite Integrals
Indefinite integrals represent the general form of antiderivatives. An indefinite integral, shown using the symbol \(\int\), signifies a family of functions that differ only by a constant. When computing an indefinite integral, we seek a function whose derivative equals the given integrand.

To solve problems involving indefinite integrals, various techniques are employed, such as substitution, integration by parts, and the use of reduction formulas. In our textbook solution case, integration by parts—a method where the integral of a product of two functions is broken down into simpler parts—is utilized. Since the constants of integration can vary, the general solution includes the symbol \(C\) to account for all possible antiderivatives.
Integral Calculus
Integral calculus is a branch of mathematics concerned with the determination of the properties and applications of integrals. It involves finding the quantities where the rate of change and the accumulation of quantities are central concepts. Applications range from computing areas and volumes to determining the behavior of physical systems.

The application of integral calculus techniques was demonstrated in the provided solution, where integration by parts was used to compute an indefinite integral of a product of polynomial and exponential functions. Understanding the interplay of different functions and utilizing various integral calculus methods, such as the reduction formulas, are essential skills for solving integration problems encountered in higher level mathematics and applied sciences.

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Most popular questions from this chapter

Evaluating an integral without the Fundamental Theorem of Calculus Evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) using the following steps. a. If \(f\) is integrable on \([0, b],\) use substitution to show that $$ \int_{0}^{b} f(x) d x=\int_{0}^{b / 2}(f(x)+f(b-x)) d x $$ b. Use part (a) and the identity \(\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\) to evaluate \(\int_{0}^{\pi / 4} \ln (1+\tan x) d x\) (Source: The College Mathematics Journal 33, 4, Sep 2004).

A differential equation of the form \(y^{\prime}(t)=F(y)\) is said to be autonomous (the function \(F\) depends only on \(y\) ). The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(F\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0,\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal line segments in the direction field. Note also that for autonomous equations, the direction field is independent of \(t\). Consider the following equations. a. Find all equilibrium solutions. b. Sketch the direction field on either side of the equilibrium solutions for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=y(y-3)$$

\(A n\) integrand with trigonometric functions in the numerator and denominator can often be converted to a rational integrand using the substitution \(u=\tan (x / 2)\) or equivalently \(x=2 \tan ^{-1} u .\) The following relations are used in making this change of variables. \(A: d x=\frac{2}{1+u^{2}} d u \quad B: \sin x=\frac{2 u}{1+u^{2}} \quad C: \cos x=\frac{1-u^{2}}{1+u^{2}}\) $$\text { Evaluate } \int_{0}^{\pi / 3} \frac{\sin \theta}{1-\sin \theta} d \theta$$

Determine whether the following statements are true and give an explanation or counterexample. a. The general solution of \(y^{\prime}(t)=20 y\) is \(y=e^{20 t}\). b. The functions \(y=2 e^{-2 t}\) and \(y=10 e^{-2 t}\) do not both satisfy the differential equation \(y^{\prime}+2 y=0\). c. The equation \(y^{\prime}(t)=t y+2 y+2 t+4\) is not separable. d. A solution of \(y^{\prime}(t)=2 \sqrt{y}\) is \(y=(t+1)^{2}\).

A total charge of \(Q\) is distributed uniformly on a line segment of length \(2 L\) along the \(y\) -axis (see figure). The \(x\) -component of the electric field at a point \((a, 0)\) is given by $$E_{x}(a)=\frac{k Q a}{2 L} \int_{-L}^{L} \frac{d y}{\left(a^{2}+y^{2}\right)^{3 / 2}}$$ where \(k\) is a physical constant and \(a>0\) a. Confirm that \(E_{x}(a)=\frac{k Q}{a \sqrt{a^{2}+L^{2}}}\) b. Letting \(\rho=Q / 2 L\) be the charge density on the line segment, show that if \(L \rightarrow \infty,\) then \(E_{x}(a)=2 k \rho / a\) (See the Guided Project Electric field integrals for a derivation of this and other similar integrals.)

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