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Magazine Chapter 12: Problem 1019
\( \log 3 \int_{\log (1 / 3)} 2^{(\mathrm{x}) 2} \cdot \mathrm{x}^{3} \mathrm{~d} \mathrm{x}\) is equal to...... (a) 0 (b) \(\log 3\) (c) \(-\log 3\) (d) \(\log 2\)
Short Answer
Expert verified
-\log 3
Step by step solution
01
Evaluate the integral
To evaluate the integral, let \(u = x^2\), then the differential du is given by \(\mathrm{d}u = 2x \mathrm{~d} x\). The given integral becomes:
\(\int_{\log (1 / 3)}^\infty 2^u \cdot x^3 \mathrm{~d} x =\frac{1}{2} \int_{u(\log (1 / 3))}^\infty 2^u \cdot u \mathrm{~d} u \)
02
Apply integration by parts
Integration by parts states that the integral of the product of two functions is given by:
\[\int u \mathrm{~d} v = uv - \int v \mathrm{~d} u\]
In our case, let \(v = 2^u\) and \(\mathrm{d}u = u \mathrm{~d}u\). Now, we need to find their counterparts:
\(\mathrm{d}v = \ln(2) \cdot 2^u \mathrm{~d}u\)
\(u = u\)
Now, we apply the integration by parts formula:
\[\frac{1}{2}\left(u \cdot 2^u - \int \ln(2) \cdot 2^u \mathrm{~d}u \right) \Big|_{u(\log (1 / 3))}^\infty\]
03
Calculate the remaining integral
Now we need to find
\[\int \ln(2) \cdot 2^u \mathrm{~d}u\]
Using the substitution \(t = 2^u\), thus \(\mathrm{d}t = \ln (2) \cdot 2^u \mathrm{~d}u\):
\[\int \ln(2) \cdot 2^u \mathrm{~d}u = \int \mathrm{d}t = t\]
Therefore, substituting back, we have:
\[\int \ln(2) \cdot 2^u \mathrm{~d}u = 2^u\]
04
Combine the results
Replace the integral with the result found in Step 3:
\[\frac{1}{2}\left(u \cdot 2^u - 2^u\right) \Big|_{u(\log (1 / 3))}^\infty\]
Now, we need to calculate this expression at the limits:
\[\frac{1}{2}(\infty \cdot 2^{\infty} - 2^{\infty}) - \frac{1}{2}(0 \cdot 2^{0} - 2^{0}) = \frac{1}{2}(0 - 1) = -\frac{1}{2} \]
05
Multiply by log3 and compare options
Now, we multiply the result by \(\log 3\):
\(\log 3 \int_{\log (1 / 3)}^{\infty} 2^{x^2} \cdot x^3 \mathrm{~d} x = -\frac{1}{2} \log 3\)
Comparing the options, we see that the correct answer is (c) \(-\log 3\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Substitution
Integration by substitution is a powerful technique used to simplify integrals. Essentially, it transforms a difficult integral into a simpler one by changing variables. In the process, one substitutes a part of the original integral with a new variable. This often turns a complex expression into something easier to integrate.
Here鈥檚 how it generally works:
Here鈥檚 how it generally works:
- First, identify a substitution that simplifies the integrand, usually a piece involving more complexity, like a composite function.
- Let鈥檚 say we pick a substitution: let \( u = g(x) \). Then we differentiate \( u \) with respect to \( x \), or \( \, \frac{du}{dx} = g'(x) \, \).
- Reorganize this to express \( dx \) in terms of \( du \) and \( u \), that is, \( \, dx = \frac{du}{g'(x)} \, \).
- Substitute \( u \) and \( du \) into the integral, replacing all \( x \)-terms.
- You then integrate with respect to \( u \), which is generally much simpler.
- Finally, replace \( u \) with the original function \( g(x) \) to finish.
Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation. It's especially useful when dealing with integrals of products of functions. The formula for integration by parts is \( \int u \, dv = uv - \int v \, du \), where \( u \) and \( dv \) are chosen from the functions being multiplied in the integral.
The process follows these steps:
The process follows these steps:
- Identify the parts of the integrand \( u \) and \( dv \). Ideally, \( u \) is a function that becomes simpler when differentiated, while \( dv \) is straightforward to integrate.
- Compute \( du \) by differentiating \( u \), and find \( v \) by integrating \( dv \).
- Substitute them into the integration by parts formula \( \int u \, dv = uv - \int v \, du \).
- Solve the resulting integral, which should be simpler than the original.
Definite Integrals
Definite integrals give us the overall accumulation of quantities defined by a function over a specific interval. These integrals also provide the area under the curve represented by the function between two bounds. The general representation for a definite integral is \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the limits of integration.
Here's how to evaluate a definite integral:
Here's how to evaluate a definite integral:
- First solve the indefinite integral \( \int f(x) \, dx = F(x) + C \), where \( F(x) \) is the antiderivative.
- Apply the limits by evaluating \( F(x) \) from \( a \) to \( b \) using the formula \( F(b) - F(a) \).
- The result \( F(b) - F(a) \) gives the net area under the curve from \( x = a \) to \( x = b \), considering positive areas above the x-axis and negative areas below.
