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Magazine Chapter 5: Problem 19
Evaluate the integrals. $$\int_{1}^{-1}(r+1)^{2} d r.$$
Short Answer
Expert verified
The value of the integral is \(-\frac{8}{3}.\)
Step by step solution
01
Identify the Integral
We need to evaluate the definite integral \[ \int_{1}^{-1}(r+1)^{2} \, dr. \]Notice that the limits of integration are reversed; usually, the lower limit is less than the upper limit.
02
Reverse the Limits of Integration
Reversing the limits of integration changes the sign of the integral. So we have:\[ \int_{1}^{-1}(r+1)^{2} \, dr = - \int_{-1}^{1}(r+1)^{2} \, dr. \]
03
Expand the Integrand
Expand the expression \( (r+1)^2 \) to make the integration easier:\[ (r+1)^2 = r^2 + 2r + 1. \]
04
Integrate the Expanded Expression
Integrate each term separately:\[\int (r^2 + 2r + 1) \, dr = \int r^2 \, dr + \int 2r \, dr + \int 1 \, dr. \]Compute each integral:- \( \int r^2 \, dr = \frac{r^3}{3} \)- \( \int 2r \, dr = 2 \cdot \frac{r^2}{2} = r^2 \)- \( \int 1 \, dr = r \)Thus, the indefinite integral is:\[ \frac{r^3}{3} + r^2 + r. \]
05
Evaluate the Definite Integral
Substitute the limits of integration into the integrated function:\[ \left[\frac{r^3}{3} + r^2 + r \right]_{-1}^{1}. \]First, evaluate at \( r = 1 \):\[ \frac{1^3}{3} + 1^2 + 1 = \frac{1}{3} + 1 + 1 = \frac{7}{3}. \]Next, evaluate at \( r = -1 \):\[ \frac{(-1)^3}{3} + (-1)^2 + (-1) = -\frac{1}{3} + 1 - 1 = -\frac{1}{3}. \]
06
Compute the Difference
Find the difference between the evaluations at the upper and lower limits:\[ \left( \frac{7}{3} \right) - \left( -\frac{1}{3} \right). \]Compute this difference:\[ \frac{7}{3} + \frac{1}{3} = \frac{8}{3}. \]Remember, from Step 2, we need to reverse the sign:\[ -\frac{8}{3}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus focuses on the concept of integrals, primarily involved in the method of finding the antiderivative of a function.
This technique essentially reverses the process of differentiation, where you find the derivative of a function.
An integral can give you the area under a curve represented by a function or solve complex problems in physics and engineering.
In our problem, we deal with a definite integral, particularly focusing on \((r+1)^2 \, dr\),
which involves using a straightforward approach to the evaluation.
This technique essentially reverses the process of differentiation, where you find the derivative of a function.
An integral can give you the area under a curve represented by a function or solve complex problems in physics and engineering.
- Definite Integrals: These are integrals with specific upper and lower limits, providing a numerical value representing the area under a curve between these limits.
- Indefinite Integrals: These do not have specific bounds and represent a family of functions rather than a single value.
In our problem, we deal with a definite integral, particularly focusing on \((r+1)^2 \, dr\),
which involves using a straightforward approach to the evaluation.
Calculus Techniques
To solve integrals effectively, several calculus techniques can be employed, including substitution, integration by parts, and partial fractions, among others.
For the given problem, the technique of expanding the integrand is used.
In our example, the limits were initially inverted and therefore reversing them was a necessary step.
Reversing limits ensures that the lower limit is less than the upper limit, soothing the process.
This expansion makes the integration straightforward by reducing complex polynomials into simpler terms: \(r^2 + 2r + 1\).
Each term can then be integrated easily on its own.
For the given problem, the technique of expanding the integrand is used.
Reversing Limits
Reversing the integration limits is a calculus technique that alters the sign of the integral.In our example, the limits were initially inverted and therefore reversing them was a necessary step.
Reversing limits ensures that the lower limit is less than the upper limit, soothing the process.
Expanding the Integrand
After reversing the limits, the function \(f(r) = (r+1)^2\) is expanded to simplify the integration.This expansion makes the integration straightforward by reducing complex polynomials into simpler terms: \(r^2 + 2r + 1\).
Each term can then be integrated easily on its own.
Evaluation of Integrals
Evaluating definite integrals involves substituting the limits of integration into the antiderivative obtained after integration.
This process helps in calculating the exact area under the curve between the given limits.
This process helps in calculating the exact area under the curve between the given limits.
- Substitution of Limits: In our example, the antiderivative obtained is \(\frac{r^3}{3} + r^2 + r\).
- Calculating Function Values: Calculate the value of the function at the upper limit (1) and lower limit (-1). It involves simple substitution of these values into the antiderivative,
yielding specific results that guide us to the complete evaluation. - Finding the Difference: The final step involves finding the difference between the upper and lower limit evaluations,
then adjusting the sign according to prior steps (like reversing limits), resulting in the evaluated integral value.
