I'm not sure what I'm doing wrong. I have attached the assignment and the work I've done. **Problem Statement:**Evaluate the following integral, where $ R $ is bounded by $ x = 1 $, $ y = 2x + 4 $, and $ y = -x - 2 $.\[\iint_R y^2 \, dA\]**Boundary Conditions Explanation:**The region $ R $ is defined by the intersection of three lines in the Cartesian plane:- The vertical line $ x = 1 $- The line $ y = 2x + 4 $, which is a straight line with a slope of 2 and a y-intercept at +4 - The line $ y = -x - 2 $, which is a straight line with a slope of -1 and a y-intercept at -2These lines form a bounded region $ R $, and the integral specifies the need to evaluate the function $ y^2 $ over this area.**Graphical Representation:**The enclosed region can be visualized by plotting these lines:1. **$ x = 1 $**: A vertical line crossing the x-axis at 1.2. **$ y = 2x + 4 $**: This line slants upwards, intersecting the y-axis at +4.3. **$ y = -x - 2 $**: This line slants downwards, intersecting the y-axis at -2.The region $ R $ is where these lines intersect each other, forming a closed area on the graph over which the double integral is evaluated to find the total contribution of $ y^2 $ within this boundary. Certainly! Here is a transcription of the text found in the image, formatted as if for an educational website:---## Double Integral Calculation**Problem** Evaluate the double integral $\int \int_R y^3 \, dA$, where $x = 1, y = 2x+4, y = -x-2$.**Given Region** - $-2 \leq x \leq 1$- $-2 \leq y \leq 2x+4$**Step-by-Step Solution**1. **Set Up Integral:** \[ \int_{-2}^{1} \left( \int_{-x-2}^{2x+4} y^2 \, dy \right) \, dx \]2. **Integrate with respect to $y$:** \[ = \int_{-2}^{1} \frac{y^3}{3} \bigg|_{-x-2}^{2x+4} \, dx \]3. **Evaluate the inner integral:** \[ = \int_{-2}^{1} \left[ \frac{(2x+4)^3}{3} - \frac{(-x-2)^3}{3} \right] \, dx \]4. **Substitute and Simplify:** \[ = \int_{-2}^{1} \frac{1}{3} \left[ 22(x+2)^3 - (-1)(x+1)^3 \right] \, dx \]5. **Change of Variables:** Let $u = x+3$, $du = dx$.6. **New Integral Bounds:** \[ = \int_{1}^{4} \frac{9}{3}(x+3)^3 \, dx \]7. **Integration with respect to $x$:** \[ = \frac{9}{3} \left[ \frac{(x+3)^4}{4} \right]_1^4 \]8. **Calculate the Definite Integral:** \[ = \frac{9}{3} \left[ \frac{64}{4} - \frac{1}{4} \

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Answered: Evaluate the following integral, where…

Math

Calculus

Evaluate the following integral, where R is bounded by x = 1, y = 2x + 4, and y=-x-2. SS₂² dA R

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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