Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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2. Evaluate the limit
![### Definite Integral Problem
Evaluate the integral below:
\[ \int_{1}^{2} (4x^3 - 3x^2 + 2x) \, dx \]
In this problem, we need to find the definite integral of the polynomial \(4x^3 - 3x^2 + 2x\) from \(x = 1\) to \(x = 2\).
To solve this, we will use the fundamental theorem of calculus, which states that if \(F(x)\) is the antiderivative of \(f(x)\), then:
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
First, we find the antiderivative of the given function \(4x^3 - 3x^2 + 2x\).
The antiderivative of \(4x^3\) is \(x^4\).
The antiderivative of \(-3x^2\) is \(-x^3\).
The antiderivative of \(2x\) is \(x^2\).
Combining these results, we have:
\[ F(x) = x^4 - x^3 + x^2 \]
Next, we apply the limits of integration:
\[ \int_{1}^{2} (4x^3 - 3x^2 + 2x) \, dx = F(2) - F(1) \]
Substituting the values, we get:
\[ F(2) = 2^4 - 2^3 + 2^2 = 16 - 8 + 4 = 12 \]
\[ F(1) = 1^4 - 1^3 + 1^2 = 1 - 1 + 1 = 1 \]
Thus,
\[ \int_{1}^{2} (4x^3 - 3x^2 + 2x) \, dx = 12 - 1 = 11 \]
So, the value of the integral is:
\[ \boxed{11} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4e699669-58d4-4b70-bb83-6285c2150cf2%2Ff23efb60-b98b-4f53-876c-bd29b20ba2b4%2Fu5jz9q_processed.png&w=3840&q=75)
Transcribed Image Text:### Definite Integral Problem
Evaluate the integral below:
\[ \int_{1}^{2} (4x^3 - 3x^2 + 2x) \, dx \]
In this problem, we need to find the definite integral of the polynomial \(4x^3 - 3x^2 + 2x\) from \(x = 1\) to \(x = 2\).
To solve this, we will use the fundamental theorem of calculus, which states that if \(F(x)\) is the antiderivative of \(f(x)\), then:
\[ \int_{a}^{b} f(x) \, dx = F(b) - F(a) \]
First, we find the antiderivative of the given function \(4x^3 - 3x^2 + 2x\).
The antiderivative of \(4x^3\) is \(x^4\).
The antiderivative of \(-3x^2\) is \(-x^3\).
The antiderivative of \(2x\) is \(x^2\).
Combining these results, we have:
\[ F(x) = x^4 - x^3 + x^2 \]
Next, we apply the limits of integration:
\[ \int_{1}^{2} (4x^3 - 3x^2 + 2x) \, dx = F(2) - F(1) \]
Substituting the values, we get:
\[ F(2) = 2^4 - 2^3 + 2^2 = 16 - 8 + 4 = 12 \]
\[ F(1) = 1^4 - 1^3 + 1^2 = 1 - 1 + 1 = 1 \]
Thus,
\[ \int_{1}^{2} (4x^3 - 3x^2 + 2x) \, dx = 12 - 1 = 11 \]
So, the value of the integral is:
\[ \boxed{11} \]
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