Chapter 5: Problem 8
Evaluate the integrals in Exercises \(1-26\) $$ \int_{-2}^{-1} \frac{2}{x^{2}} d x $$
Short Answer
Expert verified
The integral evaluates to 1.
Step by step solution
01
Understand the Integral
We need to evaluate the definite integral of \( \frac{2}{x^2} \) from \( x = -2 \) to \( x = -1 \). The function \( \frac{2}{x^2} \) is continuous over the interval \([-2, -1]\), so the Fundamental Theorem of Calculus can be applied.
02
Find the Antiderivative
First, we find the antiderivative of \( \frac{2}{x^2} \). Rewrite \( \frac{2}{x^2} \) as \( 2x^{-2} \). The antiderivative of \( 2x^{-2} \) is \( -2x^{-1} = -\frac{2}{x} \). We found this by using the power rule which states the integral of \( x^n \) is \( \frac{x^{n+1}}{n+1} \) for \( n eq -1 \).
03
Apply the Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus tells us to evaluate \( F(b) - F(a) \) where \( F \) is the antiderivative of our function. Thus, evaluate the antiderivative from \(-2\) to \(-1\):\[ -\frac{2}{-1} - \left(-\frac{2}{-2} \right) \].
04
Compute the Values
Substitute the limits of integration into the antiderivative:- At \( x = -1 \), \(-\frac{2}{-1} = 2\).- At \( x = -2 \), \(-\frac{2}{-2} = 1\).Thus, the definite integral becomes \(2 - 1 = 1\).
05
Conclude the Evaluation
The evaluation of the definite integral \( \int_{-2}^{-1} \frac{2}{x^2} \, dx \) results in \(1\). This means the area under the curve from \(-2\) to \(-1\) is 1.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus (FTC) is a powerful principle that connects differentiation and integration.The beauty of FTC lies in how it allows us to solve complex area problems with ease.There are two main parts: the first part guarantees that the derivative of an integral function gives back the original function.The second part, which we focus on here, provides a way to evaluate definite integrals efficiently.To use the FTC for definite integrals, you need:
So, thanks to this theorem, you are spared the effort of finding limits of Riemann sums.
- An integrable function over a closed interval, like our function \( \frac{2}{x^2} \) over \([-2, -1]\).
- The antiderivative \( F \) (primitive), so that \( F' = f \).
So, thanks to this theorem, you are spared the effort of finding limits of Riemann sums.
Antiderivatives
Antiderivatives, sometimes called indefinite integrals, are the reverse of derivatives.They answer the question: "What function results in our given function when differentiated?"When handling the problem \(\int \frac{2}{x^2} dx\), we seek a function whose derivative gives back \(\frac{2}{x^2} \).Rewriting it as \(2x^{-2}\) helps visualize the pattern and align it with rules we know.The antiderivative, for our case, turned out to be \(-\frac{2}{x}\).Key things to remember:
- Antiderivatives introduce a constant, often written as \(C\), representing any constant shift won't affect the derivative.
- Multiple approaches exist in finding them, but often require recognizing the form your function takes, like power, exponential, or trigonometric.
- Checking your antiderivative by differentiation can always confirm accuracy!
- In definite integrals, the constant \(C\) cancels out, simplifying the process!
Power Rule in Integrals
The power rule simplifies integration, just as it does differentiation.When applied, it gives a straightforward way to find antiderivatives for functions like \(x^n\), where \(n eq -1\).For our exercise, \(\frac{2}{x^2} \) turns into \(2x^{-2}\).The power rule then informs us that the antiderivative is \(\frac{x^{n+1}}{n+1}\).Steps involved:
When \(n = -1\), the antiderivative becomes logarithmic, resulting in \(\ln |x|\), not a power rule scenario.
Thus, the power rule elegantly and efficiently dealt with our given function, computing the antiderivative as \(-\frac{2}{x}\).
- Adjust the exponent: Increase \(n\) by 1.
- Divide the variable by the new exponent value \(n+1\).
- Apply the constant multiple rule—this means multiplying the whole expression by any coefficient outside the integral, like our 2.
When \(n = -1\), the antiderivative becomes logarithmic, resulting in \(\ln |x|\), not a power rule scenario.
Thus, the power rule elegantly and efficiently dealt with our given function, computing the antiderivative as \(-\frac{2}{x}\).