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Chapter 7: Problem 5

Using the Fundamental Theorem, evaluate the definite integrals in problem exactly. $$ \int_{1}^{2} \frac{1}{t^{2}} d t $$

Short Answer

Expert verified
The integral evaluates to \( \frac{1}{2} \).

Step by step solution

01

Identify the Function

The given integral \[ \int_{1}^{2} \frac{1}{t^{2}} \, dt \]involves the function \( f(t) = \frac{1}{t^{2}} \), which can be rewritten as \( f(t) = t^{-2} \) for easier integration.
02

Determine the Antiderivative

Find the antiderivative of \( f(t) = t^{-2} \) using the power rule for integration: \[ \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \] Applying this to \( t^{-2} \), we have: \[ \int t^{-2} \, dt = \frac{t^{-1}}{-1} = -t^{-1} = -\frac{1}{t} + C \].
03

Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, if \( F(t) \) is an antiderivative of \( f(t) \), then:\[ \int_{a}^{b} f(t) \, dt = F(b) - F(a) \] In our case, \( F(t) = -\frac{1}{t} \), so the definite integral becomes:\[ \left[-\frac{1}{t}\right]_{1}^{2} = \left(-\frac{1}{2}\right) - \left(-\frac{1}{1}\right) \].
04

Evaluate the Definite Integral

Substitute the upper and lower bounds into the antiderivative:- For \( t = 2 \): \( F(2) = -\frac{1}{2} \)- For \( t = 1 \): \( F(1) = -\frac{1}{1} = -1 \)The result is:\[ \left(-\frac{1}{2}\right) - \left(-1\right) = -\frac{1}{2} + 1 = \frac{1}{2} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Definite Integral
A definite integral involves finding the accumulated area under a curve within a specified interval. In this exercise, the given integral is \ \( \int_{1}^{2} \frac{1}{t^{2}} \, dt \ \).
  • The limits of integration, 1 and 2, define the interval.
  • The curve of interest is represented by the function \( f(t) = \frac{1}{t^{2}} \).
The integral calculates the total sum of infinitesimally small rectangles under the curve from \( t = 1 \) to \( t = 2 \).
  • This process gives us the exact area between the function and the horizontal axis within the specified bounds.
  • The Fundamental Theorem of Calculus connects the integral with an antiderivative.
The calculation shows this value to be \( \frac{1}{2} \). By evaluating the definite integral, we use the upper and lower limits to find this accumulated area.
Antiderivative
An antiderivative is crucial for solving a definite integral as it reverses the differentiation process. For this problem, we needed to find the antiderivative of the function \( f(t) = t^{-2} \).
  • The antiderivative provides a function \( F(t) \) whose derivative is \( f(t) \).
  • In this case, \( F(t) = -\frac{1}{t} \).
Once we have the antiderivative, we can use it to evaluate the definite integral.During this process:
  • Calculate \( F(b) \) and \( F(a) \), wherein \( b \) and \( a \) are the upper and lower limits of the integral, respectively.
  • Find the difference \( F(b) - F(a) \) to determine the total area.
By identifying the correct antiderivative, the problem simplifies to substituting the limits and finding the result, which in this case is \( \frac{1}{2} \).
Power Rule for Integration
The power rule for integration is an essential method used to find antiderivatives, allowing us to integrate polynomial functions.For our exercise, the function \( t^{-2} \) was dealt with using the power rule:
  • It states \( \int t^n \, dt = \frac{t^{n+1}}{n+1} + C \), where \( n eq -1 \).
  • Applying it to \( t^{-2} \), we increase the exponent by one, resulting in \( \frac{t^{-1}}{-1} \).
  • This simplifies to \( -\frac{1}{t} + C \).
Using the power rule simplifies finding antiderivatives of such functions:
  • This is invaluable because it provides a systematic approach to integration.
  • Especially useful for polynomials and similar forms.
Understanding and applying the power rule effectively enables us to solve integrals with ease, such as achieving our final result \( \frac{1}{2} \) for this exercise.

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