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Magazine Chapter 6: Problem 1
\(\int_{-1}^{1} x^{2}-x-1 d x=\) (A) \(\frac{2}{3}\) (B) 0 (C) \(-\frac{4}{3}\) (D) -2
Short Answer
Expert verified
The integral evaluates to \(-\frac{4}{3}\), option (C).
Step by step solution
01
Understand the Problem
We need to evaluate the definite integral \( \int_{-1}^{1} (x^2 - x - 1) \, dx \). The integral is a function that gives the area under the curve of \( f(x) = x^2 - x - 1 \) between \( x = -1 \) and \( x = 1 \).
02
Calculate the Indefinite Integral
First, find the antiderivative (indefinite integral) of \( x^2 - x - 1 \). Apply the power rule: \[ \int (x^2 - x - 1) \, dx = \frac{x^3}{3} - \frac{x^2}{2} - x + C \] where \( C \) is the constant of integration.
03
Apply the Fundamental Theorem of Calculus
Use the Fundamental Theorem of Calculus, which states that to evaluate the definite integral \( \int_{a}^{b} f(x) \, dx \), compute \( F(b) - F(a) \) where \( F(x) \) is an antiderivative of \( f(x) \). Here, \( a = -1 \) and \( b = 1 \).
04
Evaluate at Upper Limit
Substitute \( x = 1 \) into the antiderivative: \[ F(1) = \frac{1^3}{3} - \frac{1^2}{2} - 1 = \frac{1}{3} - \frac{1}{2} - 1 \]Simplify this expression: \[ F(1) = \frac{1}{3} - \frac{3}{6} - \frac{6}{6} = \frac{1}{3} - \frac{1}{2} - 1 \] \[ = \frac{2}{6} - \frac{3}{6} - \frac{6}{6} = \frac{-7}{6} \].
05
Evaluate at Lower Limit
Substitute \( x = -1 \) into the antiderivative: \[ F(-1) = \frac{(-1)^3}{3} - \frac{(-1)^2}{2} - (-1) \] which simplifies to: \[ F(-1) = -\frac{1}{3} - \frac{1}{2} + 1 \] \[ = -\frac{2}{6} - \frac{3}{6} + \frac{6}{6} = \frac{1}{6} \].
06
Calculate the Definite Integral
Subtract the result from Step 5 from the result of Step 4 to find the definite integral:\[ F(1) - F(-1) = \frac{-7}{6} - \frac{1}{6} = \frac{-8}{6} = -\frac{4}{3} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indefinite Integral
The indefinite integral is a fundamental concept in calculus that involves finding the antiderivative of a function. It is represented by the integral symbol \( \int \) without any upper or lower limits. This process involves calculating a function whose derivative gives back the original function. The result includes a constant of integration, represented as \( C \), which captures all possible vertical shifts of the antiderivative.
For example, when we have a function like \( x^2 - x - 1 \), we calculate its indefinite integral as:
For example, when we have a function like \( x^2 - x - 1 \), we calculate its indefinite integral as:
- Apply the power rule to each term: \( \int x^2 \, dx = \frac{x^3}{3} \)
- Next, \( \int x \, dx = \frac{x^2}{2} \)
- Lastly, \( \int 1 \, dx = x \)
- Combine these results to get the full antiderivative: \( \frac{x^3}{3} - \frac{x^2}{2} - x + C \)
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a keystone in understanding the relationship between derivatives and integrals. It essentially bridges the concepts of differentiation and integration.
Here's how it works:
Here's how it works:
- The first part states that if a function \( f(x) \) is continuous over an interval \([a, b]\) and \( F(x) \) is its antiderivative, then the definite integral of \( f(x) \) from \( a \) to \( b \) is \( F(b) - F(a) \).
- This means, instead of calculating the area under the curve through summation, we can evaluate \( F(x) \) at the boundaries and subtract: \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).
Power Rule
The power rule is an essential tool for finding antiderivatives. It helps simplify expressions involving polynomials by providing a straightforward method for integration.
When applying the power rule:
When applying the power rule:
- Considering any function \( x^n \), its antiderivative is \( \frac{x^{n+1}}{n+1} \) where \( n eq -1 \).
- This means if you have \( x^2 \), its antiderivative becomes \( \frac{x^3}{3} \).
- Similarly, for \( x \), \( n = 1 \), making the antiderivative \( \frac{x^2}{2} \).
Antiderivative Calculation
Calculating antiderivatives involves finding a function whose derivative returns the original function. This calculation is crucial for integrating functions and is particularly useful in solving definite integrals.
Here's how to proceed with antiderivative calculation:
Here's how to proceed with antiderivative calculation:
- Identify each term in the polynomial and apply the appropriate rule - often the power rule.
- Compute individually: for instance, integrating \( x^2 \), \( x \), and constants with the power rule.
- Combine these results to form the complete antiderivative, including the constant \( C \).
