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Evaluate the following integrals. $$\int t e^{t} d t$$

Short Answer

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Question: Evaluate the integral ∫ te^t dt. Answer: ∫ te^t dt = te^t - e^t + C

Step by step solution

01

Identify u and dv

Let's choose our u and dv. Here, we choose: $$u = t \qquad dv = e^t dt$$
02

Find du and v

Now let's find du and v: $$du = dt \qquad v = \int e^t dt$$ By integrating e^t, we find v: $$v = e^t$$
03

Apply integration by parts formula

Now, we apply the integration by parts formula: $$\int t e^t dt = uv - \int v du$$ Substitute the values of u, dv, du, and v that we found previously: $$\int t e^t dt = (t)(e^t) - \int (e^t)(dt)$$
04

Evaluate the remaining integral

Evaluate the remaining one-variable integral: $$\int t e^t dt = t e^t - \int e^t dt$$ Integrating the remaining part, we get: $$\int t e^t dt = t e^t - e^t + C$$ Where C is the constant of integration.

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Key Concepts

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

Indefinite Integral
An indefinite integral, also known as an antiderivative, represents the collection of all the possible antiderivatives of a given function. For instance, when you see the symbol \(\int f(x) dx\), this symbol \(\int\) is the integral sign, and the function \(f(x)\) inside the integration is the integrand. The \(dx\) at the end signals that the integration is with respect to variable \(x\).

Finding an indefinite integral is like reversing the process of differentiation. If differentiation gives you the speed at which a car moves when you know its position, then integrating the speed function gives you an equation for the position of the car. An essential feature of the indefinite integral is the constant of integration, \(C\). This constant accounts for all the possible vertical shifts of the antiderivative since differentiation of any constant is zero.

In the exercise provided, \(\int t e^{t} dt\), we are looking for a function whose derivative with respect to \(t\) is \(t e^{t}\). This is a standard type of problem in calculus known as finding the indefinite integral, or antiderivative, of a function.
Exponential Functions
Exponential functions are a fundamental class of functions characterized by an equation of the form \(f(x) = a^{x}\), where \(a\) is a positive real number not equal to 1, and \(x\) is the exponent. These functions are unique as their rate of growth (or decay if \(0 < a < 1\)) is proportional to their current value. This property leads to their frequent appearance in real-world applications ranging from compound interest to the population dynamics of species.

In the context of integration, the exponential function \(e^x\), where \(e\) (approximately 2.71828) is the base of the natural logarithm, is particularly important. It has the special property that its derivative is the same as the function itself: \(\frac{d}{dx} e^x = e^x\). This makes it relatively straightforward to integrate and differentiate compared to other functions.

In the exercise \(\int t e^{t} dt\), the presence of the exponential function \(e^{t}\) is what guides the choice of the function \(dv\) during the integration by parts process, as it remains the same upon differentiation, which simplifies the calculation.
Integration Techniques
Integration techniques are various methods used to calculate definite and indefinite integrals. Some common techniques include substitution, integration by parts, partial fractions, and trigonometric substitution. Each technique is useful for different kinds of functions and scenarios, and a well-equipped mathematician will be familiar with multiple methods to tackle various integration problems.

Integration by parts is one of the primary methods used when the integrand is a product of two functions that cannot easily be integrated together. The formula, based on the product rule for differentiation, is given by \(\int u dv = uv - \int v du\). This method requires you to cleverly choose \(u\) and \(dv\) such that \(du\) and \(v\) are easier to work with.

The exercise \(\int t e^{t} dt\) is a classic example where integration by parts is the appropriate technique. By choosing \(u = t\) and \(dv = e^{t} dt\), we take advantage of both the simplicity of differentiating \(t\) and the invariant property of the exponential function upon integrating \(e^{t}\), leading to a solution with manageable terms. This strategic choice is a crucial skill for successfully applying integration techniques.

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Most popular questions from this chapter

Approximate the following integrals using Simpson's Rule. Experiment with values of \(n\) to ensure that the error is less than \(10^{-3}\). \(\int_{0}^{\pi} \sin 6 x \cos 3 x d x=\frac{4}{9}\)

An endowment is an investment account in which the balance ideally remains constant and withdrawals are made on the interest earned by the account. Such an account may be modeled by the initial value problem \(B^{\prime}(t)=a B-m\) for \(t \geq 0,\) with \(B(0)=B_{0} .\) The constant \(a\) reflects the annual interest rate, \(m\) is the annual rate of withdrawal, and \(B_{0}\) is the initial balance in the account. a. Solve the initial value problem with a=0.05, m= 1000 dollar \(/\mathrm{yr}\), and \(B_{0}\)= 15,000 dollar. Does the balance in the account increase or decrease? b. If \(a=0.05\) and \(B_{0}\)= 50,000 dollar, what is the annual withdrawal rate \(m\) that ensures a constant balance in the account? What is the constant balance?

Another Simpson's Rule formula is \(S(2 n)=\frac{2 M(n)+T(n)}{3},\) for \(n \geq 1 .\) Use this rule to estimate \(\int_{1}^{e} 1 / x d x\) using \(n=10\) subintervals.

Bob and Bruce bake bagels (shaped like tori). They both make bagels that have an inner radius of 0.5 in and an outer radius of 2.5 in. Bob plans to increase the volume of his bagels by decreasing the inner radius by \(20 \%\) (leaving the outer radius unchanged). Bruce plans to increase the volume of his bagels by increasing the outer radius by \(20 \%\) (leaving the inner radius unchanged). Whose new bagels will have the greater volume? Does this result depend on the size of the original bagels? Explain.

Compute \(\int_{0}^{1} \ln x d x\) using integration by parts. Then explain why \(-\int_{0}^{\infty} e^{-x} d x\) (an easier integral) gives the same result.

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