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Chapter 4: Problem 27

find the derivative \(f^{\prime}(x)\) \(f(x)=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\)

Short Answer

Expert verified
The derivative of the function \(f(x)=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) is \(f^{\prime}(x) = x^{2}-3x+2\)

Step by step solution

01

Understand the problem

Given the function \(f(x)=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\). We need to find the derivative of this function \(f^{\prime}(x)\). We need to apply the fundamental theorem of calculus to evaluate this.
02

Apply the Fundamental Theorem of calculus

The fundamental theorem of calculus (part 1) tells us that if a function \(f\) is continuous on the closed interval \([a, b]\) and \(F\) is the antiderivative of \(f\) on the interval, then the definite integral of \(f\) from \(a\) to \(b\) is equal to \(F(b) - F(a)\). However, in our case, \(b=x\), therefore according to the fundamental theorem of calculus, the derivative of our function \(f(x)\) will be the function that we are integrating evaluated at \(x\). That is, \(f^{\prime}(x) = x^{2}-3x+2\)
03

Write out the result

So, the derivative of the function \(f(x)=\int_{0}^{x}\left(t^{2}-3 t+2\right) d t\) is \(f^{\prime}(x) = x^{2}-3x+2\)

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Key Concepts

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

Derivative of a function
The derivative of a function represents the rate at which the function's value changes with respect to a change in the input value. In simpler terms, it tells us how fast or slow something is happening at any given point.
  • A function's derivative can help identify points where it reaches a maximum or minimum.
  • It also helps in determining the slope of a curve, which is crucial in graphing.
  • Derivatives are used extensively in physics to determine velocity and acceleration.
In the context of our problem, we have a function defined as an integral, and by using the Fundamental Theorem of Calculus, we easily find its derivative. This theorem greatly simplifies our task. For our function, the derivative is the integrand evaluated at the upper limit, simplifying the process of differentiation for such functions.
Continuous function
A continuous function is a function that does not have any breaks, jumps, or holes in its graph. Essentially, you can draw it without lifting your pencil from the paper.
  • Continuity is crucial for applying theorems in calculus, such as the Fundamental Theorem of Calculus.
  • A function must be continuous on an interval to ensure that its integral can be evaluated smoothly over that interval.
  • Without continuity, calculus operations can become unpredictable or undefined.
In our given problem, the integrand function, which is the polynomial \(t^2 - 3t + 2\), is continuous everywhere. Therefore, it meets the criteria needed to apply our calculus theorem without any trouble.
Definite integral
A definite integral is used to calculate the area under a curve between two specific points, often providing useful information about accumulations or net changes.
  • The result of a definite integral is a number, unlike an indefinite integral which is a function.
  • Definite integrals have limits of integration, which define the region over which the area is calculated.
  • They are used in various fields such as physics and engineering to calculate displacement, area, volume, and more.
In our exercise, we deal with a definite integral from 0 to \(x\), describing how the area under the polynomial \(t^2 - 3t + 2\) changes as \(x\) varies. Using the Fundamental Theorem of Calculus, the process to find the function's behavior as \(x\) changes becomes straightforward.
Antiderivative
An antiderivative of a function is a function whose derivative is the original function. Finding an antiderivative is effectively reversing the process of differentiation.
  • Every continuous function has an infinite number of antiderivatives.
  • The antiderivative includes a constant of integration, representing family of functions differing by a constant.
  • They are crucial for solving integral problems and determining overall behavior of functions over intervals.
In our given exercise, identifying that the definite integral itself stems from an antiderivative function helps us apply the Fundamental Theorem of Calculus effectively. We rely on understanding that the derivative of the function \(F(x)\) yields the original function we are integrating, reaffirming our solution's correctness as it satisfied the theorem's conditions.

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Most popular questions from this chapter

Find the average value of the function on the given interval. \(f(x)=\cos x,[0, \pi / 2]\)

Evaluate the integral. $$\int_{0}^{1} \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} d x$$

Evaluate the integral exactly, if possible. Otherwise, estimate it numerically. (a) \(\int_{0}^{\pi} \sin x^{2} d x\) (b) \(\int_{0}^{\pi} x \sin x^{2} d x\)

Evaluate the integral. $$\int_{e}^{e^{2}} \frac{1}{x \ln x} d x$$

In most of the calculations that you have done, it is true that the Trapezoidal Rule and Midpoint Rule are on opposite sides of the exact integral (i.e., one is too large, the other too small). Also, you may have noticed that the Trapezoidal Rule tends to be about twice as far from the exact value as the Midpoint Rule.Given this, explain why the linear combination \(\frac{1}{3} T_{n}+\frac{2}{3} M_{n}\) should give a good estimate of the integral. (Here, \(T_{n}\) represents the Trapezoidal Rule approximation using \(n\) partitions and \(M_{n}\) the corresponding Midpoint Rule approximation.)
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