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Magazine Chapter 6: Problem 44
Evaluate each definite integral using integration by parts. (Leave answers in exact form.) \(\int_{0}^{4} 2(z-4)^{6} d z\)
Short Answer
Expert verified
The definite integral evaluates to \(-\frac{147456}{7}\).
Step by step solution
01
Choose 'u' and 'dv' for Integration by Parts
Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int 2(z-4)^6 \, dz \), we choose \( u = (z-4)^6 \) and \( dv = 2 \, dz \).
02
Differentiate and Integrate Terms
Differentiate \( u \) to get \( du = 6(z-4)^5 \, dz \), and integrate \( dv = 2 \, dz \) to find \( v = 2z \).
03
Apply the Integration by Parts Formula
Substituting into the integration by parts formula, we have: \( \int 2(z-4)^6 \ dz = (z-4)^6 (2z) \bigg| _0^4 - \int 2z \times 6(z-4)^5 \ dz \).
04
Calculate the Boundary Terms
Evaluate \((z-4)^6 (2z)\) at the boundaries. At \(z = 4\), it is \(0\) because \((4-4)^6 = 0\). At \(z = 0\), it simplifies to \((-4)^6 \times 0 = 0\). Therefore, the boundary terms contribute 0 to the integral.
05
Simplify the Remaining Integral
We focus on the integral \(- \int 12z(z-4)^5 \, dz\). Recognize the power reduction by substituting: \( x = z-4 \), \( dx = dz\), changing the limits from \(z=0\) to \(x=-4\) and \(z=4\) to \(x=0\), resulting in \(-12 \int (x+4)x^5 \, dx\).
06
Solution Through Simple Polynomial Integration
Expand and integrate: \(-12 \left( \int x^6 \, dx + 4 \int x^5 \, dx \right) = -12 \left( \frac{x^7}{7} + 4 \frac{x^6}{6} \right) + C\), where the integration bounds are from -4 to 0.
07
Calculate the Definite Integral Result
Evaluate the expressions at the bounds after integration: \(-12 \left( 0-\left(\frac{-4^7}{7} + \frac{4}{6}(-4)^6 \right) \right)\). Simplify to get \(-12 \left( -\frac{16384}{7} + 4096\right)\).
08
Final Simplification
Simplify the evaluation: \(= -12 (-\frac{16384}{7} + \frac{28672}{7}) = -12 \frac{12288}{7} = -\frac{147456}{7}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Let's dive into what makes a definite integral unique in calculus. A definite integral calculates the area under a curve from one point to another. It's a way of summing infinitely many tiny areas to find the total area. This is different from an indefinite integral, which doesn't have set limits and results in a family of functions.
In our problem, we are dealing with the integral from 0 to 4 of the expression \(2(z-4)^6\). The limits, 0 to 4, are crucial because they define the starting and ending points for finding the area under the curve. These bounds tell us exactly which part of the graph we are interested in.
Calculating a definite integral using integration by parts lets us better manage complex functions. But keep in mind, after simplifying with other techniques, you always have to plug in these boundary values to find the solution.
In our problem, we are dealing with the integral from 0 to 4 of the expression \(2(z-4)^6\). The limits, 0 to 4, are crucial because they define the starting and ending points for finding the area under the curve. These bounds tell us exactly which part of the graph we are interested in.
Calculating a definite integral using integration by parts lets us better manage complex functions. But keep in mind, after simplifying with other techniques, you always have to plug in these boundary values to find the solution.
Integration Techniques
Integration by parts is an incredibly handy technique in calculus, especially for integrals involving products of functions. Inspired by the product rule from differentiation, it helps break down complicated products into more manageable integrals.
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Choosing \(u\) and \(dv\) effectively is crucial for simplifying the integral. In our example, we have \(u = (z-4)^6\) and \(dv = 2 \, dz\). Then, we need to differentiate \(u\) and integrate \(dv\) to continue.
The magic happens when applying this technique because you transform a tough integral into a combination of simpler parts. Always keep in mind the integration technique is only successful if the resulting integrals are easier to solve than the original.
The formula for integration by parts is:
\[ \int u \, dv = uv - \int v \, du \]
Choosing \(u\) and \(dv\) effectively is crucial for simplifying the integral. In our example, we have \(u = (z-4)^6\) and \(dv = 2 \, dz\). Then, we need to differentiate \(u\) and integrate \(dv\) to continue.
The magic happens when applying this technique because you transform a tough integral into a combination of simpler parts. Always keep in mind the integration technique is only successful if the resulting integrals are easier to solve than the original.
- Differentiate \(u\) to get \(du\).
- Integrate \(dv\) to obtain \(v\).
- Substitute and simplify using the integration by parts formula.
Calculus Problem Solving
Calculus problem solving can seem daunting, but breaking it down into steps helps. First, understand the problem by identifying what you're solving for, which in this case was a complex integral between limits.
Next, choose the most appropriate method or technique. For our integral \(\int_0^4 2(z-4)^6 \, dz\), integration by parts was the chosen method because of the polynomial form, which suits breaking into parts.
After setting up, compute each part carefully to avoid mistakes. This includes differentiating and integrating as needed, simplifying the integral as much as possible. Recognize any substitutions that might ease calculations, like changing variables to align better with limits, as we did from \(z\) to \(x = z-4\).
Next, choose the most appropriate method or technique. For our integral \(\int_0^4 2(z-4)^6 \, dz\), integration by parts was the chosen method because of the polynomial form, which suits breaking into parts.
After setting up, compute each part carefully to avoid mistakes. This includes differentiating and integrating as needed, simplifying the integral as much as possible. Recognize any substitutions that might ease calculations, like changing variables to align better with limits, as we did from \(z\) to \(x = z-4\).
- Set clear goals for each part of the problem.
- Break down complex functions using the right integration or substitution technique.
- After calculation, always double-check by plugging boundary values in definite integrals.
