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Magazine Chapter 7: Problem 29
Evaluate the following integrals. $$\int_{-1}^{1} \frac{x}{(x+3)^{2}} d x$$
Short Answer
Expert verified
Question: Evaluate the definite integral $$\int_{-1}^{1} \frac{x}{(x+3)^{2}}d x$$
Answer: $$\int_{-1}^{1} \frac{x}{(x+3)^{2}}d x = \ln 2 + \frac{3}{4}$$
Step by step solution
01
Identify substitution
Let's choose a substitution that simplifies the denominator. We have:
$$u = x+3$$
Now find the derivative and solve for dx:
$$\frac{d u}{d x} = 1 \Longrightarrow d u = d x$$
02
Rewrite the integral
Substituting \(u\) and \(du\) into the given integral, we get:
$$\int_{-1}^{1} \frac{x}{(x+3)^{2}} d x = \int_{(-1+3)}^{(1+3)} \frac{u-3}{u^2}\,du$$
Note the limits change according to our substitution, \(u(-1)=(-1+3)=2\) and \(u(1)=(1+3)=4\).
03
Break down the integral and simplify
Divide the integral into two parts and simplify:
$$\int_{2}^{4} \frac{u-3}{u^2}\,du = \int_{2}^{4} \frac{u}{u^2} - \frac{3}{u^2}\,du = \int_{2}^{4} \frac{1}{u} - 3u^{-2} \, du$$
04
Evaluate the antiderivatives
Now find the antiderivatives of the terms in the integral:
$$\int{ \frac{1}{u}}\,du = \ln u + C_1$$
$$\int{3u^{-2}}\,du = -3u^{-1} + C_2$$
So, the antiderivative of the integral is:
$$F(u) = \ln u -3u^{-1} + C$$
05
Apply the Fundamental Theorem of Calculus
Now, using the definite integral limits, we have:
$$\int_{2}^{4} \frac{1}{u} - 3u^{-2} \, du = F(4)-F(2) = [\ln 4 - \frac{3}{4}] - [\ln 2 - \frac{3}{2}] $$
06
Compute the result
Compute the final result:
$$[\ln 4 - \frac{3}{4}] - [\ln 2 - \frac{3}{2}] = \ln \frac{4}{2} - \frac{3}{4} +\frac{3}{2} = \ln 2 + \frac{3}{2} -\frac{3}{4} = \ln 2 + \frac{3}{4} $$
Thus, the final result is: $$\int_{-1}^{1} \frac{x}{(x+3)^{2}}d x = \ln 2 + \frac{3}{4}$$
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integrals
Definite integrals are an essential part of calculus used to find the area under a curve, within specified limits. They are different from indefinite integrals, which represent a family of functions. A definite integral provides a numerical value, offering a precise measurement of the total accumulation between two points on the x-axis.
For example, in the exercise we evaluated, the definite integral \( \int_{-1}^{1} \frac{x}{(x+3)^{2}} dx \) finds the area under the curve of the function \( \frac{x}{(x+3)^{2}} \) from \( x = -1 \) to \( x = 1 \).
The resulting value \( \ln 2 + \frac{3}{4} \) gives a clear, numerical representation of this area.
For example, in the exercise we evaluated, the definite integral \( \int_{-1}^{1} \frac{x}{(x+3)^{2}} dx \) finds the area under the curve of the function \( \frac{x}{(x+3)^{2}} \) from \( x = -1 \) to \( x = 1 \).
The resulting value \( \ln 2 + \frac{3}{4} \) gives a clear, numerical representation of this area.
- Limits of Integration: The bounds of the integral, here \(-1\) to \(1\), are essential in determining the interval over which the function is integrated.
- Significance: The result, a single numerical value, helps quantify various physical concepts like distance, area, and even electrical charges.
Antiderivatives
Antiderivatives, often referred to as the "reverse process" of differentiation, are used to find a function whose derivative is the given function. They are a crucial step in the process of integration, transforming complex expressions into more manageable forms.
In the solution provided, the antiderivative was used to break the function into simpler parts, specifically \( \frac{1}{u} \) and \(-3u^{-2} \).
These were then integrated separately to find their respective antiderivatives \( \ln u \) and \(-3u^{-1} \).
In the solution provided, the antiderivative was used to break the function into simpler parts, specifically \( \frac{1}{u} \) and \(-3u^{-2} \).
These were then integrated separately to find their respective antiderivatives \( \ln u \) and \(-3u^{-1} \).
- Purpose: By finding the antiderivatives, we simplify the computation of definite integrals.
- Process: Each part of the split integral is integrated, enhancing the understanding of each term's contribution to the overall value.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus links the concept of differentiation and integration, two core operations in calculus. It provides a unifying framework that underlines the connection between finding antiderivatives and evaluating definite integrals.
In our example, once the antiderivative \( F(u) = \ln u - 3u^{-1} + C \) was found, the theorem was applied to evaluate the definite integral from 2 to 4.
This application leads to calculating \( F(4) - F(2) \), linking the antiderivative to a specific number: \( \ln 2 + \frac{3}{4} \).
In our example, once the antiderivative \( F(u) = \ln u - 3u^{-1} + C \) was found, the theorem was applied to evaluate the definite integral from 2 to 4.
This application leads to calculating \( F(4) - F(2) \), linking the antiderivative to a specific number: \( \ln 2 + \frac{3}{4} \).
- Part 1: It states that if \( F \) is an antiderivative of \( f \) on an interval, then \( \int_{a}^{b} f(x) dx = F(b) - F(a) \).
- Part 2: It allows us to compute antiderivatives conveniently, thus finding areas and other quantities when combined with limits.
