/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 In each of Exercises \(21-28,\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In each of Exercises \(21-28,\) calculate the derivative of \(F(x)\) with respect to \(x\). $$ F(x)=\int_{0}^{x} \tan ^{2}(t) d t $$

Short Answer

Expert verified
The derivative is \(F'(x) = \tan^2(x)\).

Step by step solution

01

Recognize the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus Part 1 states that if \(F(x) = \int_{a}^{x} f(t) \, dt\), then the derivative \(F'(x) = f(x)\). We need to identify \(f(t)\).
02

Identify the Function for the Derivative

In our given integral, \(f(t) = \tan^2(t)\). Based on the Fundamental Theorem of Calculus, differentiate \(F(x) = \int_{0}^{x} \tan^2(t) \, dt\) to find \(F'(x)\).
03

Apply the Fundamental Theorem of Calculus

According to the theorem, the derivative of \(F(x)\) is simply the integrand function evaluated at \(x\), hence \(F'(x) = \tan^2(x)\).
04

Verify the Result

Review the computation to ensure that applying the theorem directly gives the derivative of \(F(x)\). Since \(f(t) = \tan^2(t)\) is continuous at all \(t\), \(F'(x) = \tan^2(x)\) is correct.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Integrals and Derivatives
In the world of calculus, integrals and derivatives are like two sides of the same coin. They are tied together by the Fundamental Theorem of Calculus, which bridges the gap between these concepts. To start, an integral can be thought of as the accumulation of values over a range for a given function. This process helps us find the total area under a curve. On the other hand, a derivative focuses on finding the rate at which something changes.

With the Fundamental Theorem of Calculus, we learn something powerful. It tells us that the derivative of an integral of a function from a constant to a variable (say from 0 to x) is just the original function evaluated at that variable. In simple terms, you can think of it as the derivative undoing what the integral did, returning us to where we started: the original function.
  • Step 1: Start with an integral of a function.
  • Step 2: Take the derivative with respect to the upper limit.
  • Step 3: The result is the function itself evaluated at the upper limit.
In our exercise, we begin with an integral of \( an^2(t)\) from 0 to \(x\). Applying this theorem, we quickly arrive at its derivative being simply \(\tan^2(x)\). It's the art of working backwards and reveals the interplay between accumulation (integral) and instantaneous change (derivative).
Continuous Functions
A continuous function is one without any breaks, jumps, or gaps in its graph. Continuity is crucial for applying the Fundamental Theorem of Calculus because it guarantees that the integral and derivative operations behave nicely.

If a function is continuous on a closed interval, then integrating or differentiating it won't lead to any unexpected behaviors. This ensures that techniques like integration and differentiation work seamlessly. In our exercise, the function \(\tan^2(t)\) is continuous for all real numbers. Such continuity affirms that applying the theorem will give us a valid derivative.
  • Provides a smooth and unbroken process for integration.
  • Ensures derivatives at all points are predictable.
  • Makes sure values do not suddenly spike to infinity or drop to negative infinity within the range of interest.
The confidence in calculating \(F'(x)\) as \(\tan^2(x)\) relies on this foundational property of continuity. With continuity, we know that using calculus techniques won't give us biased or incorrect results.
Differentiation
Differentiation is both a fundamental process and an art form in calculus. It involves finding the derivative of a function, which illustrates the function's rate of change at any given point. Think of it like looking at the speedometer on a car that tells you how fast you're going at any given second.

The process of differentiation uses formulas and rules—developed over centuries of mathematical exploration—to simplify the process of finding the derivative. When the Fundamental Theorem of Calculus is applied, differentiation becomes straightforward and, at times, almost instantaneous.
  • Identify the function you want to differentiate.
  • Apply differentiation rules or theorems where applicable.
  • Simplify the result if necessary.
In the exercise with \(F(x) = \int_{0}^{x} \tan^2(t) \, dt\), differentiation through the theorem allows us to leap directly to \(F'(x) = \tan^2(x)\). This shortcut highlights the elegance and efficiency of calculus when differentiating a definite integral. Differentiation is not only for finding instantaneous change but also for unlocking the tangible relationship between variables in mathematical functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The graphs of \(y=f(x)\) and \(y=g(x)\) intersect in more than two points. Find the total area of the regions that are bounded above and below by the graphs of \(f\) and \(g\). $$ f(x)=2 \sin (x) \quad g(x)=6(\pi-x) /(5 \pi) $$

The integral \(\int_{a}^{b}\left(f_{1}(x)-f_{2}(x)\right) d x\) represents the area of a region in the \(x y\) -plane that is bounded by the graphs of \(f_{1}\) and \(f_{2}\). Express the area of the region as an integral of the form \(\int_{c}^{d}\left(g_{1}(y)-g_{2}(y)\right) d y .\) For example, the integral \(\int_{0}^{1}\left(x-x^{2}\right) d x\) represents the area of the shaded region in Figure \(11 .\) This area can also be represented as \(\int_{0}^{1}(\sqrt{y}-y) d y.\) $$ \int_{0}^{4}(\sqrt{x}-x / 2) d x $$

A function \(f\) is defined on a specified interval \(I=[a, b] .\) Calculate the area of the region that lies between the vertical lines \(x=a\) and \(x=b\) and between the graph of \(f\) and the \(x\) -axis. $$ f(x)=\exp (x)-e \quad I=[-1,2] $$

An integral \(\int_{a}^{b} f(x) d x\) and a positive integer \(N\) are given. Compute the exact value of the integral, the Simpson's Rule approximation of order \(N,\) and the absolute error \(\varepsilon\). Then find a value \(c\) in the interval \((a, b)\) such that \(\varepsilon=(b-a)^{5}\left|f^{(4)}(c)\right| /\left(180 \cdot N^{4}\right) .\) (This form of the error, which resembles the Mean Value Theorem, implies inequality \((5.8 .4) .)\) $$ \int_{1}^{4} \sqrt{x} d x \quad N=6 $$

Express the area of the given region as a sum of integrals of the form \(\int_{a}^{b} f(x) d x\). The triangle with vertices (1,0),(3,1),(2,2)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.