/*! 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 6 Evaluate the integrals in Exerci... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Evaluate the integrals in Exercises \(1-28\). $$\int_{-2}^{3}\left(x^{3}-2 x+3\right) d x$$

Short Answer

Expert verified
The integral evaluates to \(\frac{105}{4}\).

Step by step solution

01

Understanding the Problem

We need to evaluate the definite integral of the function \(f(x) = x^3 - 2x + 3\) from \(x = -2\) to \(x = 3\). This can be done by finding the antiderivative of the function and applying the Fundamental Theorem of Calculus.
02

Find the Antiderivative

The first step in evaluating the integral is to find the antiderivative of the function \(x^3 - 2x + 3\). The antiderivative of \(x^3\) is \(\frac{x^4}{4}\), for \(-2x\) it is \(-x^2\), and for constant 3 it is \(3x\). Thus, the antiderivative is \(F(x) = \frac{x^4}{4} - x^2 + 3x\).
03

Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that if \(F\) is the antiderivative of \(f\), then \(\int_{a}^{b} f(x)\,dx = F(b) - F(a)\). So evaluate \(F(3)\) and \(F(-2)\).
04

Evaluate \(F(3)\)

Substitute \(x = 3\) into the antiderivative: \(F(3) = \frac{3^4}{4} - (3)^2 + 3(3) = \frac{81}{4} - 9 + 9 = \frac{81}{4}\).
05

Evaluate \(F(-2)\)

Substitute \(x = -2\) into the antiderivative: \(F(-2) = \frac{(-2)^4}{4} - (-2)^2 + 3(-2) = \frac{16}{4} - 4 - 6 = 4 - 4 - 6 = -6\).
06

Final Calculation

Subtract \(F(-2)\) from \(F(3)\): \(\int_{-2}^{3} (x^3 - 2x + 3) dx = \frac{81}{4} - (-6) = \frac{81}{4} + 6 = \frac{81}{4} + \frac{24}{4} = \frac{81 + 24}{4} = \frac{105}{4}\).

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.

Antiderivative
The antiderivative of a function is like its inverse operation. Think of it as the reverse of differentiation. If we have a function, say, a polynomial like \( x^3 \), the antiderivative helps us find a new function whose derivative is exactly \( x^3 \). This is crucial when we want to calculate integrals, especially when dealing with definite integrals. The antiderivative of a polynomial involves increasing the power of each term by one and then dividing by this new power.

For example, finding the antiderivative of \( x^3 \) involves these steps:
  • Increase the power: from \( x^3 \) to \( x^4 \).
  • Divide by the new power: the antiderivative becomes \( \frac{x^4}{4} \).
Repeating this process for each term in a polynomial gives us the entire antiderivative. Evaluating a definite integral becomes straightforward once we find this antiderivative.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus connects differentiation and integration, serving as a bridge between these two fundamental concepts in calculus. It tells us how to evaluate definite integrals if we know the antiderivative of a function. Here's how it works: If \( F \) is the antiderivative of \( f \), then the definite integral over an interval \([a, b]\) is given by \[ \int_{a}^{b} f(x)\,dx = F(b) - F(a). \]

This theorem simplifies the computation of definite integrals tremendously. Instead of approximating the area under the curve directly, we can focus on the easier task of finding an antiderivative. Once found, substituting the upper limit and lower limit into the antiderivative gives us the result. This theorem not only simplifies calculations but also highlights the elegant interplay between the slopes of curves and the areas beneath those curves.
Polynomial Integration
Polynomial integration is the process of finding the integral of polynomial functions. Polynomials are a series of terms where each term is a variable raised to a power, multiplied by a coefficient. Integrating a polynomial is straightforward because it involves applying the power rule of antiderivatives to each term. For instance, to integrate \( x^3 - 2x + 3 \), we treat each term individually:

  • \( x^3 \) becomes \( \frac{x^4}{4} \).
  • \(-2x \) becomes \(-x^2 \).
  • The constant \(3\) becomes \(3x\).
Combining these results gives us the complete antiderivative, \( F(x) = \frac{x^4}{4} - x^2 + 3x \). This makes it easy to compute the definite integral, as shown in the exercise. Polynomial functions have straightforward integration rules, aiding in many calculus calculations and applications.

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

If av(f) really is a typical value of the integrable function \(f(x)\) on \([a, b],\) then the constant function av(f) should have the same integral over \([a, b]\) as \(f .\) Does it? That is, does $$ \int_{a}^{b} \operatorname{av}(f) d x=\int_{a}^{b} f(x) d x ? $$ Give reasons for your answer.

Each of the following functions solves one of the initial value problems in Exercises \(55-58 .\) Which function solves which problem? Give brief reasons for your answers. \begin{equation} \begin{array}{ll}{\text { a. } y} & {=\int_{1}^{x} \frac{1}{t} d t-3 \quad\quad \text { b. } y=\int_{0}^{x} \sec t d t+4} \\ {\text { c. } y} & {=\int_{-1}^{x} \sec t d t+4 \quad \text { d. } y=\int_{\pi}^{x} \frac{1}{t} d t-3}\end{array} \end{equation} $$y^{\prime}=\sec x, \quad y(0)=4$$

In Exercises \(91-94,\) you will find the area between curves in the plane when you cannot find their points of intersection using simple algebra. Use a CAS to perform the following steps: a. Plot the curves together to see what they look like and how many points of intersection they have. b. Use the numerical equation solver in your CAS to find all the points of intersection. c. Integrate \(|f(x)-g(x)|\) over consecutive pairs of intersection values. d. Sum together the integrals found in part (c). $$ f(x)=\frac{x^{4}}{2}-3 x^{3}+10, \quad g(x)=8-12 x $$

Another proof of the Evaluation Theorem \begin{equation} \begin{array}{l}{\text { a. Let } a=x_{0} < x_{1} < x_{2} \cdots < x_{n}=b \text { be any partition of }[a, b],} \\ {\text { and let } F \text { be any antiderivative of } f \text { . Show that }}\end{array} \end{equation} \begin{equation} F(b)-F(a)=\sum_{i=1}^{n}\left[F\left(x_{i}\right)-F\left(x_{i-1}\right)\right]. \end{equation} \begin{equation} \begin{array}{l}{\text { b. Apply the Mean Value Theorem to each term to show that }} \\\ {F\left(x_{i}\right)-F\left(x_{i-1}\right)=f\left(c_{i}\right)\left(x_{i}-x_{i-1}\right) \text { for some } c_{i} \text { in the interval }} \\ {\left(x_{i-1}, x_{i}\right) . \text { Then show that } F(b)-F(a) \text { is a Riemann sum for } f} \\ {\text { on }[a, b] .} \\ {\text { c. From part (b) and the definition of the definite integral, show }} \\ {\text { that }}\end{array} \end{equation} \begin{equation} F(b)-F(a)=\int_{a}^{b} f(x) d x. \end{equation}

Find the area of the propeller-shaped region enclosed by the curves \(x-y^{1 / 3}=0\) and \(x-y^{1 / 5}=0\)

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.