Use integration to find the area between the functions y = x² and y = 3x + 4. Show your work. -4 YA 20 15+ 10+ 5+ O -5- g(x) = 3x + 4 Sybas and fix) = x² 204 6 X
Use integration to find the area between the functions y = x² and y = 3x + 4. Show your work. -4 YA 20 15+ 10+ 5+ O -5- g(x) = 3x + 4 Sybas and fix) = x² 204 6 X
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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![### Finding the Area Between Curves Using Integration
#### Problem Statement:
Use integration to find the area between the functions \( y = x^2 \) and \( y = 3x + 4 \). Show your work.
#### Solution:
Firstly, identify the two functions and set them equal to find the points of intersection:
\[ x^2 = 3x + 4 \]
Rearrange to form a quadratic equation:
\[ x^2 - 3x - 4 = 0 \]
Factorize the quadratic equation:
\[ (x - 4)(x + 1) = 0 \]
Solving for \( x \), we get:
\[ x = 4 \quad \text{and} \quad x = -1 \]
These values are the points of intersection and will be our limits of integration.
The area between the curves is given by the integral of the difference between the upper function \( g(x) = 3x + 4 \) and the lower function \( f(x) = x^2 \) from \( x = -1 \) to \( x = 4 \):
\[ \text{Area} = \int_{-1}^{4} [(3x + 4) - x^2] \, dx \]
\[ \text{Area} = \int_{-1}^{4} (3x + 4 - x^2) \, dx \]
Integrate term by term:
\[ \int (3x + 4 - x^2) \, dx = \int 3x \, dx + \int 4 \, dx - \int x^2 \, dx \]
\[ = \left[ \frac{3x^2}{2} \right]_{-1}^{4} + \left[ 4x \right]_{-1}^{4} - \left[ \frac{x^3}{3} \right]_{-1}^{4} \]
Evaluate each term at the upper and lower bounds:
\[ \left( \frac{3(4)^2}{2} + 4(4) - \frac{(4)^3}{3} \right) - \left( \frac{3(-1)^2}{2} + 4(-1) - \frac{(-1)^3}{3} \right) \]
Calculate each value:
\[ \left(](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cd02c50-b0b4-4862-9c83-ad8fe712a454%2F75ea914f-d49d-4ac3-ad6d-6bad92bfed94%2Fm623i1i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Area Between Curves Using Integration
#### Problem Statement:
Use integration to find the area between the functions \( y = x^2 \) and \( y = 3x + 4 \). Show your work.
#### Solution:
Firstly, identify the two functions and set them equal to find the points of intersection:
\[ x^2 = 3x + 4 \]
Rearrange to form a quadratic equation:
\[ x^2 - 3x - 4 = 0 \]
Factorize the quadratic equation:
\[ (x - 4)(x + 1) = 0 \]
Solving for \( x \), we get:
\[ x = 4 \quad \text{and} \quad x = -1 \]
These values are the points of intersection and will be our limits of integration.
The area between the curves is given by the integral of the difference between the upper function \( g(x) = 3x + 4 \) and the lower function \( f(x) = x^2 \) from \( x = -1 \) to \( x = 4 \):
\[ \text{Area} = \int_{-1}^{4} [(3x + 4) - x^2] \, dx \]
\[ \text{Area} = \int_{-1}^{4} (3x + 4 - x^2) \, dx \]
Integrate term by term:
\[ \int (3x + 4 - x^2) \, dx = \int 3x \, dx + \int 4 \, dx - \int x^2 \, dx \]
\[ = \left[ \frac{3x^2}{2} \right]_{-1}^{4} + \left[ 4x \right]_{-1}^{4} - \left[ \frac{x^3}{3} \right]_{-1}^{4} \]
Evaluate each term at the upper and lower bounds:
\[ \left( \frac{3(4)^2}{2} + 4(4) - \frac{(4)^3}{3} \right) - \left( \frac{3(-1)^2}{2} + 4(-1) - \frac{(-1)^3}{3} \right) \]
Calculate each value:
\[ \left(
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