Problem 8 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-26\) $$ \int_{-2}^{-1} \frac{2}{x^{2}} d x $$

Step-by-Step Solution

Answer

The integral evaluates to 1.

Step 1: 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.

Step 2: 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 \( n eq -1 \).

Step 3: 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 \(-1\):\[ -\frac{2}{-1} - \left(-\frac{2}{-2} \right) \].

Step 4: 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 \(2 - 1 = 1\).

Step 5: 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.

Key Concepts

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:

• An integrable function over a closed interval, like our function \( \frac{2}{x^2} \) over \([-2, -1]\).
• The antiderivative \( F \) (primitive), so that \( F' = f \).

You then compute the difference \( F(b) - F(a) \), where \( a \) and \( b \) are the bounds.In essence, it swiftly computes the net area under the curve between the two points!
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:

• 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.

But why not \(n = -1\)?
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 \(-\frac{2}{x}\).