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Magazine Chapter 5: Problem 14
Let \(f(t)=F^{\prime}(t) .\) Write the integral \(\int_{a}^{b} f(t) d t\) and evaluate it using the Fundamental Theorem of Calculus. $$F(t)=\tan t ; a=0, b=\pi$$
Short Answer
Expert verified
The value of the integral is 0.
Step by step solution
01
Identify f(t) and F(t)
Given the problem statement, we know that \( f(t) = F'(t) \). Since \( F(t) = \tan t \), it follows that \( f(t) = \frac{d}{dt} \tan t = \sec^2 t \).
02
Set up the Integral
We need to evaluate the definite integral \( \int_{0}^{\pi} f(t) \, dt \). From Step 1, we substitute \( f(t) \) with \( \sec^2 t \) to get \( \int_{0}^{\pi} \sec^2 t \, dt \).
03
Apply the Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \( \int_{a}^{b} f(t) \, dt = F(b) - F(a) \). Here \( F(t) = \tan t \), so we evaluate \( F(b) - F(a) \) as \( \tan(\pi) - \tan(0) \).
04
Compute the Values
Calculate \( \tan(\pi) \) and \( \tan(0) \). Since \( \tan(\pi) = 0 \) and \( \tan(0) = 0 \), we find that \( \tan(\pi) - \tan(0) = 0 - 0 = 0 \).
05
Conclude the Evaluation
The value of the integral \( \int_{0}^{\pi} \sec^2 t \, dt \) is 0, as found from the previous step.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite integral
The concept of a definite integral is widely used in calculus to find the total accumulation of a quantity over an interval. It is expressed as \( \int_{a}^{b} f(t) \, dt \), which essentially means the area under the curve of function \( f(t) \) from a point \( a \) to a point \( b \).
This calculation is crucial when determining things like total distance traveled, area under a curve, or total accumulation of any variable rate process over a given time or space.
This calculation is crucial when determining things like total distance traveled, area under a curve, or total accumulation of any variable rate process over a given time or space.
- In the given problem, the focus is on finding \( \int_{0}^{\pi} \sec^2 t \, dt \).
- The role of this definite integral, in this case, is to evaluate this total area under the function \( \sec^2 t \) from 0 to \( \pi \).
Antiderivative
Finding an antiderivative is central to solving a definite integral using the Fundamental Theorem of Calculus.
An antiderivative is essentially a function \( F(t) \) whose derivative \( F'(t) \) gives us the original function \( f(t) \).
An antiderivative is essentially a function \( F(t) \) whose derivative \( F'(t) \) gives us the original function \( f(t) \).
- In our problem, \( F(t) = \tan t \) serves as the antiderivative of \( f(t) = \sec^2 t \).
- This means that \( \frac{d}{dt} \tan t = \sec^2 t \), verifying that \( \tan t \) is indeed the antiderivative of \( \sec^2 t \).
Tangent function
The tangent function, represented as \( \tan t \), is a fundamental trigonometric function used to relate angles to side ratios in a right triangle.
It is expressed as the ratio of the opposite side to the adjacent side in the context of a right triangle.
It is expressed as the ratio of the opposite side to the adjacent side in the context of a right triangle.
- In calculus, the tangent function not only has geometric importance but also significant applications in slope and rate calculation because it relates to the angle's slope.
- In this exercise, \( \tan t \) serves as the antiderivative to \( \sec^2 t \).
Derivative of tangent
Understanding the derivative of the tangent function is crucial as it reveals the rate at which the tangent function changes.
The derivative of \( \tan t \) is \( \sec^2 t \), and this is directly used as the function \( f(t) \) in the calculus problem we are examining.
The derivative of \( \tan t \) is \( \sec^2 t \), and this is directly used as the function \( f(t) \) in the calculus problem we are examining.
- This derivative process demonstrates how rates of change can be identified, with \( \sec^2 t \) representing the rate of change of \( \tan t \).
- Calculus allows us to connect these functions through derivatives and antiderivatives, establishing a relationship between them to solve definite integrals.
