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Magazine Chapter 14: Problem 7
Evaluate the integrals using integration by parts where possible. $$\int\left(x^{2}+1\right) e^{-2 x+4} d x$$
Short Answer
Expert verified
The evaluated integral using integration by parts is \(\int (x^{2}+1) e^{-2x+4} dx = -\frac{1}{2}x^2e^{-2x+4} -\frac{1}{2}xe^{-2x+4} -\frac{3}{4}e^{-2x+4} + C\).
Step by step solution
01
Identify u and dv
For integration by parts, let's select our functions u and dv. Here, we'll choose \(u = x^2\) and \(dv = e^{-2x+4}dx\).
02
Find du and v
Now, we need to find the derivatives of the u function and the integral of the dv function, which are:
\(du = 2x dx\),
\(v = \int e^{-2x+4} dx\).
To find the integral of v, we have:
\(v = \int e^{-2x+4} dx = -\frac{1}{2}e^{-2x+4}+C\).
03
Apply integration by parts to x^2 term
Now, using the integration by parts formula: \(\int u dv = uv - \int v du\), we apply it to the \(x^2e^{-2x+4}\) term:
\(\int x^2 e^{-2x+4} dx = x^2(-\frac{1}{2}e^{-2x+4}) - \int (-\frac{1}{2}e^{-2x+4})(2x) dx\).
Simplify the equation:
\(\int x^2 e^{-2x+4} dx = -\frac{1}{2}x^2e^{-2x+4} + \int xe^{-2x+4} dx\).
Now, we'll apply integration by parts again to \(\int xe^{-2x+4} dx\) with \(u = x\) and \(dv = e^{-2x+4} dx\).
04
Find du and v for second integration by parts
We find the derivative of the u function and the integral of the dv function:
\(du = dx\),
\(v = -\frac{1}{2}e^{-2x+4}\).
05
Apply integration by parts to x term
Applying the integration by parts formula again to the (\(x e^{-2x+4}\)) term:
\(\int xe^{-2x+4} dx = x(-\frac{1}{2}e^{-2x+4}) - \int(-\frac{1}{2} e^{-2x+4}) dx\).
Simplify the equation:
\(\int xe^{-2x+4} dx = -\frac{1}{2}xe^{-2x+4} + \frac{1}{2}\int e^{-2x+4} dx\).
Using the same integral we calculated for \(v\) from Step 2, we can substitute it back into the equation:
\(\int xe^{-2x+4} dx = -\frac{1}{2}xe^{-2x+4} -\frac{1}{4}e^{-2x+4} + C\).
06
Integrate the remaining term
Now, integrate the \((1 \cdot e^{-2x+4})\) term:
\(\int e^{-2x+4} dx = -\frac{1}{2}e^{-2x+4}+C\).
07
Add the results and simplify
Combining the results from steps 3, 5, and 6, we get:
\(\int (x^{2}+1) e^{-2x+4} dx = -\frac{1}{2}x^2e^{-2x+4} -\frac{1}{2}xe^{-2x+4} -\frac{1}{4}e^{-2x+4} -\frac{1}{2}e^{-2x+4} + C\).
Simplify the equation:
\(\int (x^{2}+1) e^{-2x+4} dx = -\frac{1}{2}x^2e^{-2x+4} -\frac{1}{2}xe^{-2x+4} -\frac{3}{4}e^{-2x+4} + C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indefinite Integral
The indefinite integral, often written as \( \int f(x)dx \), represents the antiderivative of a function. This is a fundamental concept in calculus, which not only reverses the process of differentiation but also helps in finding the area under a curve. Unlike a definite integral, an indefinite integral does not have bounds and includes a constant of integration (\(C\)) because the antiderivative of a function is not unique.
When evaluating an indefinite integral, the goal is to find a function whose derivative would produce the original function under the integral sign. This is where knowledge of derivatives comes into play. For instance, the integral of the exponential function \(e^x\) is \(e^x + C\), since the derivative of \(e^x\) is itself. Understanding the relationship between a function and its derivative is crucial in determining the antiderivative.
When evaluating an indefinite integral, the goal is to find a function whose derivative would produce the original function under the integral sign. This is where knowledge of derivatives comes into play. For instance, the integral of the exponential function \(e^x\) is \(e^x + C\), since the derivative of \(e^x\) is itself. Understanding the relationship between a function and its derivative is crucial in determining the antiderivative.
Exponential Functions
In the context of integration, exponential functions are significant because they maintain their form after differentiation and integration. An exponential function has the form \(a^x\), where \(a\) is a positive real number, and \(x\) is the exponent. The most common base for exponential functions in calculus is the natural number \(e\), approximately equal to 2.71828.
Exponential functions such as \(e^x\) and its scaled variants like \(e^{ax+b}\) appear frequently in various disciplines, including physics, finance, and engineering, owing to their unique properties. For example, they are heavily used in problems dealing with growth and decay, as they model processes that change exponentially over time. In integration, when you come across an exponential function, you often use substitution or integration by parts to evaluate it, depending on the other terms in the integral.
Exponential functions such as \(e^x\) and its scaled variants like \(e^{ax+b}\) appear frequently in various disciplines, including physics, finance, and engineering, owing to their unique properties. For example, they are heavily used in problems dealing with growth and decay, as they model processes that change exponentially over time. In integration, when you come across an exponential function, you often use substitution or integration by parts to evaluate it, depending on the other terms in the integral.
Derivatives
The concept of a derivative is central to both differential and integral calculus. It represents the rate of change of a function with respect to a variable, commonly denoted as \(\frac{df}{dx}\) or \(f'(x)\). A strong understanding of derivatives is critical in the process of integration by parts.
One of the core strategies in performing integration by parts is choosing the correct function to differentiate (\(u\)) and the correct function to integrate (\(dv\)). The general rule of thumb is to pick a \(u\) that simplifies upon differentiation, making the integrand easier to handle. In our exercise, we differentiated \(x^2\) to get \(2x\), which is simpler and thus a good choice for \(u\). Similarly, knowing that the derivative of \(e^x\) is \(e^x\) helps anticipate that integrating an exponential function will yield a function of the same form, thereby guiding the choice for \(dv\).
One of the core strategies in performing integration by parts is choosing the correct function to differentiate (\(u\)) and the correct function to integrate (\(dv\)). The general rule of thumb is to pick a \(u\) that simplifies upon differentiation, making the integrand easier to handle. In our exercise, we differentiated \(x^2\) to get \(2x\), which is simpler and thus a good choice for \(u\). Similarly, knowing that the derivative of \(e^x\) is \(e^x\) helps anticipate that integrating an exponential function will yield a function of the same form, thereby guiding the choice for \(dv\).
