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Magazine Chapter 5: Problem 53
(a) Find \(\int_{0}^{1} f(3 x+1) d x\) if \(\int_{1}^{4} f(x) d x=5\). (b) Find \(\int_{0}^{3} f(3 x) d x\) if \(\int_{0}^{9} f(x) d x=5\). (c) Find \(\int_{-2}^{0} x f\left(x^{2}\right) d x\) if \(\int_{0}^{4} f(x) d x=1\).
Short Answer
Expert verified
(a) \( \frac{5}{3} \), (b) \( \frac{5}{3} \), (c) \( \frac{1}{2} \)
Step by step solution
01
Simplify the Integral Limit for (a)
We are given \(\int_{0}^{1} f(3x+1) \, dx\), and we need to change the variable to match the given information. Let \( u = 3x + 1 \), then \( du = 3 \, dx \) or \( dx = \frac{1}{3} \, du \). When \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 4 \). Thus,\[ \int_{0}^{1} f(3x+1) \, dx = \int_{1}^{4} f(u) \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{1}^{4} f(u) \, du \]We know \( \int_{1}^{4} f(x) \, dx = 5 \), hence:\[ \int_{0}^{1} f(3x+1) \, dx = \frac{1}{3} \times 5 = \frac{5}{3} \]
02
Simplify the Integral in (b)
We are given \(\int_{0}^{3} f(3x) \, dx\). Let \( u = 3x \), then \( du = 3 \, dx \) or \( dx = \frac{1}{3} \, du \). When \( x = 0 \), \( u = 0 \) and when \( x = 3 \), \( u = 9 \). Thus,\[ \int_{0}^{3} f(3x) \, dx = \int_{0}^{9} f(u) \cdot \frac{1}{3} \, du = \frac{1}{3} \int_{0}^{9} f(u) \, du \]We know \( \int_{0}^{9} f(x) \, dx = 5 \), so:\[ \int_{0}^{3} f(3x) \, dx = \frac{1}{3} \times 5 = \frac{5}{3} \]
03
Simplify the Integral in (c) using a Change of Variable
We are given \(\int_{-2}^{0} x f(x^2) \, dx\). Use the substitution \( u = x^2 \), then \( du = 2x \, dx \) or \( x \, dx = \frac{1}{2} \, du \). When \( x = -2 \), \( u = 4 \) and when \( x = 0 \), \( u = 0 \). Therefore, the integral becomes:\[ \int_{-2}^{0} x f(x^2) \, dx = -\frac{1}{2} \int_{4}^{0} f(u) \, du = \frac{1}{2} \int_{0}^{4} f(u) \, du \](where we've accounted for the change in limits by introducing a negative sign). Given \( \int_{0}^{4} f(x) \, dx = 1 \), we have:\[ \int_{-2}^{0} x f(x^2) \, dx = \frac{1}{2} \times 1 = \frac{1}{2} \]
04
Conclusion
Each integral has been simplified using appropriate substitutions and the given integral values. Thus:(a) \( \int_{0}^{1} f(3x+1) \, dx = \frac{5}{3} \)(b) \( \int_{0}^{3} f(3x) \, dx = \frac{5}{3} \)(c) \( \int_{-2}^{0} x f(x^2) \, dx = \frac{1}{2} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding U-Substitution
U-substitution is a technique in integral calculus that makes evaluating an integral easier by simplifying it into a recognizably simpler form. It鈥檚 almost like a way to "rename" the problem to help find a solution. Here's the basic idea:
- First, we pick a part of the integral to "substitute" - this becomes our new variable, usually denoted as \( u \). For example, in the problem, substitution was used with \( u = 3x + 1 \) and \( u = 3x \).
- The next step involves differentiating this \( u \) expression. This process helps us in rewriting the differential \( dx \) in terms of \( du \). For instance, from \( u = 3x + 1 \), we get \( du = 3 \, dx \) or \( dx = \frac{1}{3} \, du \).
- Lastly, we adjust the limits of the integral to match our new variable, \( u \). This involves substituting the original limits of \( x \) into our \( u \) equation, which we did in the exercise to change \( 0 \) to \( 1 \) and \( 1 \) to \( 4 \).
Exploring Definite Integrals
Definite integrals are a fundamental concept in calculus that involve integrating a function over a specific interval. This process computes the "net area" under the curve of a function between two points, providing meaningful information about the function's behavior in that range. Here are some key points about definite integrals:
- The "definite" part means we are looking for a number, a specific value that represents the area. This contrasts with indefinite integrals, which represent a general function.
- The notation \( \int_{a}^{b} f(x) \, dx \) denotes the integral of the function \( f(x) \) as \( x \) ranges from \( a \) to \( b \).
- It's crucial to remember that definite integrals account for the direction on the \( x \)-axis. If the curve dips below the \( x \)-axis, it impacts whether the area is added or subtracted.
The Role of Change of Variables
The method called 'change of variables' is a cornerstone in simplifying complex integrals, closely related to u-substitution. In essence, this technique alters the terms of the integral to facilitate easier integration. Here's how it works:
- The idea begins with recognizing a complex component in the integral which can be substituted to make the expression simpler, changing both the variable and the limits.
- In our exercise, for problems like \( \int_{-2}^{0} x f(x^2) \, dx \), we used the substitution \( u = x^2 \) making \( du = 2x \, dx \), (so \( x \, dx = \frac{1}{2} \, du \).
- Not only the variables, but the limits of the integrals are changed by substituting the limits of the previous variable into the new variable equation.
