Question

\( \log 3 \int_{\log (1 / 3)} 2^{(\mathrm{x}) 2} \cdot \mathrm{x}^{3} \mathrm{~d} \mathrm{x}\) is equal to...... (a) 0 (b) \(\log 3\) (c) \(-\log 3\) (d) \(\log 2\)

Short answer

-\log 3

Step-by-step solution

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: \(\int_{\log (1 / 3)}^\infty 2^u \cdot x^3 \mathrm{~d} x =\frac{1}{2} \int_{u(\log (1 / 3))}^\infty 2^u \cdot u \mathrm{~d} u \)

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: \[\frac{1}{2}\left(u \cdot 2^u - \int \ln(2) \cdot 2^u \mathrm{~d}u \right) \Big|_{u(\log (1 / 3))}^\infty\]

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: \[\int \ln(2) \cdot 2^u \mathrm{~d}u = 2^u\]

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: \[\frac{1}{2}(\infty \cdot 2^{\infty} - 2^{\infty}) - \frac{1}{2}(0 \cdot 2^{0} - 2^{0}) = \frac{1}{2}(0 - 1) = -\frac{1}{2} \]

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) \(-\log 3\).

Key concepts

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’s how it generally works:

• First, identify a substitution that simplifies the integrand, usually a piece involving more complexity, like a composite function.
• Let’s 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.

In the exercise, the substitution \( u = x^2 \) transformed the integral, helping to simplify it into a more manageable form for further calculation.

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:

• 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.

In the problem you’re working with, we have \( v = 2^u \) and \( du = u \, du \), and these choices allowed us to simplify the original integral through methodical application of integration by parts, contributing to a streamlined solution.

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:

• 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.

In the original exercise, the definite integral was assessed from specific bounds, with calculations determining the result as \(-\frac{1}{2} \log 3\). This shows the contribution of the area under the function \(2^{x^2} x^3\) from the lower to the upper limit.