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Magazine Chapter 16: Problem 187
The value of the integral \(I=\int_{0}^{1} x(1-x)^{n} d x\) is (A) \(\frac{1}{n+1}\) (B) \(\frac{1}{n+2}\) (C) \(\frac{1}{n+1}-\frac{1}{n+2}\) (D) \(\frac{1}{n+1}+\frac{1}{n+2}\)
Short Answer
Expert verified
The value of the integral is \(\frac{1}{(n+1)(n+2)}\). Option C matches.
Step by step solution
01
Understand the Integral
We need to solve the integral \(I=\int_{0}^{1} x(1-x)^{n} dx\). This is a definite integral with the variable of integration being \(x\) from 0 to 1.
02
Use Integration by Parts
We apply the method of integration by parts where \(u = x\) and \(dv = (1-x)^n dx\). Therefore, \(du = dx\) and to find \(v\), we integrate \(dv\):\[v = \int (1-x)^n dx.\]
03
Integrate \(dv\)
To integrate \((1-x)^n\), we use the substitution \(t = 1-x\), hence \(dt = -dx\). The integral becomes:\[\int (1-x)^n dx = -\int t^n dt = -\frac{t^{n+1}}{n+1} + C = -\frac{(1-x)^{n+1}}{n+1}.\]
04
Apply Integration by Parts Formula
With \(u = x\), \(du = dx\), and \(v = -\frac{(1-x)^{n+1}}{n+1}\), the integration by parts formula \(\int u \, dv = uv - \int v \, du\) gives:\[I = \left[-\frac{x(1-x)^{n+1}}{n+1}\right]_{0}^{1} + \int_0^1 \frac{(1-x)^{n+1}}{n+1} dx.\]
05
Evaluate and Simplify the First Term
Evaluate \([-\frac{x(1-x)^{n+1}}{n+1}]_{0}^{1}\): At \(x = 1\), the term becomes 0, and at \(x = 0\), the term also becomes 0. Thus, the first part of the solution is 0.
06
Simplify the Second Term
The second term is simplified as:\[\int_0^1 \frac{(1-x)^{n+1}}{n+1} dx = \frac{1}{n+1}\int_0^1 (1-x)^{n+1} dx.\]Now let \(t = 1-x\), then \(dt = -dx\), changing the limits gives \(t\) from 1 to 0:\[= \frac{1}{n+1} \int_1^0 t^{n+1} (-dt) = \frac{1}{n+1} \int_0^1 t^{n+1} dt.\]
07
Integrate the Simplified Term
Continuing from the previous step, we integrate:\[= \frac{1}{n+1} \left[ \frac{t^{n+2}}{n+2} \right]_{0}^{1} = \frac{1}{(n+1)(n+2)} \times (1 - 0).\]
08
Combine Results
The portions together reveal:\[I = 0 + \frac{1}{(n+1)(n+2)}.\]Adding the above, we find the solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration by Parts
Integration by Parts is a method used to solve integrals that are products of functions. The formula is based on the product rule for differentiation and is given by:
- \(\int u \, dv = uv - \int v \, du\)
- Identifying parts of the integrand to assign to \(u\) and \(dv\) where \(dv\) is easy to integrate.
- Computing \(du\) and \(v\) from the derivatives of \(u\) and integrating \(dv\).
- Filling them into the formula to simplify the original integral.
Substitution Method
The Substitution Method is often used to simplify integrals by transforming them into a more manageable form. The steps typically involve:
- Choosing a substitution that will simplify the integral.
- Changing the variable of integration from \(x\) to a new variable, such as \(t\).
- Replacing \(dx\) with \(dt\) using the derivative of the substitution.
Mathematical Problem Solving
Mathematical Problem Solving in the context of definite integrals involves a strategic approach to break down complex problems and apply the right mathematical methods. The process usually follows these steps:
- Understanding the problem and identifying what is given and what needs to be found.
- Choosing an appropriate integration technique based on the form of the integral.
- Evaluating the integral, often by using limits to determine the specific value over an interval.
- Simplifying results to find a comprehensible solution.
