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

Short Answer

Expert verified
The evaluated integral using integration by parts is: \(\int \frac{x^{2}-x}{e^{x}} d x = - x^2 e^{-x} - xe^{-x} + 3e^{-x} + C\)

Step by step solution

01

Choose u and dv

We need to choose a function for u and another function for dv. In general, select u to be the expression that becomes simpler after taking the derivative. In our case, this would be the polynomial part. So, we have: \[u = x^2 - x\] Now, we can choose the rest of the expression (including the dx) for dv: \[dv = \frac{1}{e^x} dx\]
02

Find du and v

Now, we need to find the derivatives and antiderivatives for u and dv, respectively: \[du = (2x - 1) dx\] \[v = \int \frac{1}{e^x} dx\] To find the antiderivative of dv, we observe that \(\frac{1}{e^x} = e^{-x}\). Thus, our antiderivative is: \[v = -e^{-x}\]
03

Apply Integration by Parts

Now we have all the components to apply the integration by parts formula: \[\int u dv = uv - \int v du\] Substitute u, v, and du into the formula: \[\int (x^2 - x) \cdot \frac{1}{e^x} dx = - (x^2 - x)e^{-x} - \int (-e^{-x})(2x - 1) dx\]
04

Simplify the Integral

Now we can simplify to obtain the remaining integral that needs to be computed: \[- (x^2 - x)e^{-x} + \int (2x - 1)e^{-x} dx\] To evaluate this integral, we will apply integration by parts once more.
05

Choose u and dv (again)

This time, we will choose: \[u = 2x - 1\] \[dv = e^{-x} dx\]
06

Find du and v (again)

Now, we need to find the derivatives and antiderivatives for u and dv, respectively: \[du = 2 dx\] \[v = -e^{-x}\]
07

Apply Integration by Parts (again)

Now, apply the integration by parts formula again: \[\int u dv = uv - \int v du\] Substitute u, v, and du into the formula: \[\int (2x - 1) e^{-x} dx = -(2x-1)e^{-x} + \int (-2e^{-x}) dx\]
08

Simplify the Integral (again)

Now we can simplify the remaining integral that needs to be computed: \[-(2x-1)e^{-x} - 2\int e^{-x} dx\] The final integral can be computed directly: \[-(2x-1)e^{-x} + 2e^{-x} + C\]
09

Combine Results

Now, combine the results of the two integration by parts calculations: \[- (x^2 - x)e^{-x} + (-(2x-1)e^{-x} + 2e^{-x} + C)\] Simplify the expression to get the final answer: \[- x^2 e^{-x} + xe^{-x} - 2xe^{-x} + e^{-x} + 2e^{-x} + C\] Thus, the evaluated integral using integration by parts is: \[\nabla \int \frac{x^{2}-x}{e^{x}} d x = - x^2 e^{-x} - xe^{-x} + 3e^{-x} + C\]

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Key Concepts

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

Understanding the Indefinite Integral
An indefinite integral, often represented as \( \int f(x) dx \), is a fundamental concept in calculus that involves finding the antiderivative of a function. What this means is that you are tasked with discovering a new function, let's denote it \( F(x) \), whose derivative yields the original function, \( f(x) \). It's as if you're tracing the steps back to the origin of a river from its current position.

Integrals are vital for a plethora of applications, whether calculating the area under a curve or predicting the accumulated quantity over time. Thinking about an indefinite integral without clear boundaries is like considering the potential for accumulating infinite possibilities. It comes handy with a constant \( C \), which represents all potential vertical shifts of \( F(x) \).
The Antiderivative: Reverse Engineering Functions
The antiderivative can be thought of as the reverse operation of differentiation. While differentiation gives you the slope of a curve at any point, the antiderivative paints a whole picture of the landscape. Unearth an antiderivative, and you've essentially reverse-engineered the function that, when differentiated, returns your original function.

So, when you perform integration by parts—a technique that elegantly breaks down complex products of functions—you're putting together a puzzle. By identifying parts of your function as \( u \) and \( dv \) and then finding their respective derivatives and antiderivatives (\( du \) and \( v \) respectively), you're deconstructing to reconstruct a complete and simpler solution.
Polynomial Functions: The Building Blocks
Polynomial functions are like the building blocks or the 'ABCs' of algebra. Defined by the sum of terms like \( ax^n \), where \( a \) is a constant and \( n \) is a non-negative integer, they lay the groundwork for more complex operations in calculus, such as integration.

Polynomials are friendly entities due to their straightforward behavior. They're easy to differentiate and integrate because of their clear-cut patterns, making them ideal candidates when selecting the \( u \) in integration by parts. For example, in our exercise, \( x^2 - x \) is a polynomial that simplifies upon differentiation, turning a potentially complicated integral into a more manageable task.
Exponential Functions: Growth and Decay
Exponential functions model phenomena that change rapidly—growing or decaying at rates proportional to their size. They take the form \( e^{rt} \), where \( e \) is the mathematical constant roughly equal to 2.71828, \( r \) represents the rate, and \( t \) signifies the time.

With their unique properties, like the rate of change itself being an exponential function, they offer distinct challenges and opportunities in calculus. For instance, \( e^{-x} \) from our exercise delineates a decaying process. The beauty of e's properties is evidenced during integration, providing a smooth operation that's almost as elegant as the function itself.

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