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Magazine Chapter 8: Problem 32
Evaluate the following integrals using integration by parts. $$\int_{0}^{1} x^{2} 2^{x} d x$$
Short Answer
Expert verified
Question: Evaluate the integral: $\int_{0}^{1} x^{2} 2^{x} d x$
Answer: The integral evaluates to $\frac{1}{\ln(2)} - \frac{2}{\ln(2)^3}$.
Step by step solution
01
Choose u and dv
Set u equal to \(x^2\) and dv equal to \(2^x dx\).
02
Find du and v
Differentiate u to find du: \(du = 2x \, dx\). Integrate dv to find v: \(v = \int 2^{x} \, dx = \frac{1}{\ln(2)}2^x\).
03
Apply integration by parts
Using the formula \(\int u \, dv = uv - \int v \, du\), we have:
$$\int_{0}^{1} x^{2} 2^{x} d x = \left[x^2 \frac{1}{\ln(2)} 2^x\right]_0^1 - \int_{0}^{1} \frac{1}{\ln(2)}2^x \cdot 2x \, dx$$
04
Simplify the expression
Now, we need to simplify the integral and evaluate it:
$$\left[x^2 \frac{1}{\ln(2)} 2^{x}\right]_0^1 - \frac{2}{\ln(2)} \int_{0}^{1} x 2^{x} dx$$
05
Apply integration by parts again
Now, for the remaining integral, let \(u = x\) and \(dv = 2^x dx\), and find \(du\) and \(v\):
\(du = dx\) and \(v = \frac{1}{\ln(2)}2^x\). Apply the integration by parts formula:
$$\int_{0}^{1} x 2^{x} d x = \left[ x \frac{1}{\ln(2)} 2^x \right]_0^1 - \int_{0}^{1} \frac{1}{\ln(2)} 2^x dx$$
06
Simplify and evaluate
Now, evaluate the definite integrals and simplify:
$$\left[ x \frac{1}{\ln(2)} 2^x \right]_0^1 - \left[ \frac{1}{\ln{2}^2} 2^x\right]_0^1 = \frac{1}{\ln(2)} - \frac{1}{\ln{2}^2}(2 - 1)$$
07
Substituting back
Now, substitute this result back into the first integral:
$$\left[ x^2 \frac{1}{\ln(2)} 2^x\right]_0^1 - \frac{2}{\ln(2)}\left(\frac{1}{\ln(2)} - \frac{1}{\ln{2}^2}\right)$$
08
Final evaluation
Evaluating the expression gives the result:
$$\frac{1}{\ln(2)} - \frac{2}{\ln(2)^2}\left(\frac{1}{\ln(2)} - \frac{1}{\ln{2}^2}\right) = \frac{1}{\ln(2)} - \frac{2}{\ln(2)^3}$$
Thus,
$$\int_{0}^{1} x^{2} 2^{x} d x = \frac{1}{\ln(2)} - \frac{2}{\ln(2)^3}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals represent the net area under a curve over a given interval. In the context of our example, where we are solving \[\int_{0}^{1} x^{2} 2^{x} d x\], the definite integral is evaluated from the lower limit of 0 to the upper limit of 1. This specific integral is challenging because it combines polynomial and exponential functions, requiring advanced techniques like integration by parts to solve. The process involves subtracting the area below the x-axis from the total area when the graph dips below.
To evaluate definite integrals with accuracy, it is essential to understand the limits of integration and the function behavior within these limits. This includes knowing when and how to apply properties of definite integrals, such as linearity, additivity, and their relationship with antiderivatives.
To evaluate definite integrals with accuracy, it is essential to understand the limits of integration and the function behavior within these limits. This includes knowing when and how to apply properties of definite integrals, such as linearity, additivity, and their relationship with antiderivatives.
Exponential Functions
Exponential functions are vital in calculus due to their unique properties; they remain unchanged when differentiated or integrated. In our exercise, the function \(2^{x}\) represents an exponential function with base 2. Exponential functions grow or decay at a rate proportional to their current value, which is why they often model real-world phenomena like population growth or radioactive decay.
When integrating exponential functions, especially during integration by parts, it is crucial to identify the right part of the function to set as \(dv\) in the integration by parts formula. In our step-by-step solution, \(2^{x} dx\) is selected as \(dv\) because its integral is known and manageable, leading to a simpler integral for further steps.
When integrating exponential functions, especially during integration by parts, it is crucial to identify the right part of the function to set as \(dv\) in the integration by parts formula. In our step-by-step solution, \(2^{x} dx\) is selected as \(dv\) because its integral is known and manageable, leading to a simpler integral for further steps.
Antiderivatives
The antiderivative, also known as an indefinite integral, is the inverse process of differentiation. When finding the antiderivative of a function, you are essentially determining a function whose derivative is the original function. It's used extensively in the process of integration by parts, which is essentially the integration application of the product rule of differentiation.
In our exercise, finding the antiderivative of \(2^{x}\) required recognizing that it equated to \(\frac{1}{\ln(2)}2^x\), which then allowed for the integration by parts process to proceed. It's important to correctly compute antiderivatives to avoid errors in solving definite integrals, and these computations provide the foundation for determining areas under curves and solving differential equations.
In our exercise, finding the antiderivative of \(2^{x}\) required recognizing that it equated to \(\frac{1}{\ln(2)}2^x\), which then allowed for the integration by parts process to proceed. It's important to correctly compute antiderivatives to avoid errors in solving definite integrals, and these computations provide the foundation for determining areas under curves and solving differential equations.
