Respuesta :
Answer:
[tex]\displaystyle \int^{1}_{0} \int^{2-y}_{y^3} dx\ dy = \frac{5}{4}[/tex]
General Formulas and Concepts:
Calculus
Integration
- Integrals
Integration Rule [Reverse Power Rule]: [tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
Integration Property [Addition/Subtraction]: [tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]
Multivariable Calculus
Integration
- Integrals
Fubini's Theorem: [tex]\displaystyle \iint_R{f(x, y)} \, dA = \int\limits^b_a \int\limits^{g_2(x)}_{g_1(x)} {f(x, y)} \, dy \, dx[/tex]
- Horizontal Cross Section: [tex]\displaystyle \iint_R {f(x, y)} \ dA = \int\limits^{y = d}_{y = c} \int\limits^{x = h(y)}_{x = g(y)} {f(x, y)} \, dx \, dy[/tex]
Area of a Plane Region Formula: [tex]\displaystyle A = \iint_R dA[/tex]
Explanation:
Step 1: Define
Identify.
[tex]\displaystyle \left \{ \begin{array}{ccc}x = y^3 \\ x = 0 \\ x + y = 2 \end{array}[/tex]
See the graph of the given curves in the attachment.
Step 2: Find Area Pt. 1
- [Curves] Rearrange: [tex]\displaystyle \left \{ \begin{array}{ccc}x = y^3 \\ x = 0 \\ x = 2 - y \end{array}[/tex]
- [Graph] Define bounds of integration c and d: [tex]\displaystyle \left \{ {{d = 2} \atop {c = 0}} \right.[/tex]
- [Graph] Identify bounds of integration g(y) and h(y): [tex]\displaystyle \left \{ {{h(y) = 2 - x} \atop {g(y) = y^3}} \right.[/tex]
Step 3: Find Area Pt. 2
- [Area of a Plane Region Formula] Derive [Fubini's Theorem]: [tex]\displaystyle A = \iint_R dA = \int\limits^b_a \int\limits^{g_2(x)}_{g_1(x)} {} \, dy \, dx[/tex]
- [Area] Derive [Horizontal Cross Section]: [tex]\displaystyle A = \int\limits^b_a \int\limits^{g_2(x)}_{g_1(x)} {} \, dy \, dx = \int\limits^{y = d}_{y = c} \int\limits^{x = h(y)}_{x = g(y)} {} \, dx \, dy[/tex]
- [Area] Substitute in variables: [tex]\displaystyle A = \int\limits^{1}_{0} \int\limits^{2 - y}_{y^3} {} \, dx \, dy[/tex]
- [Inner Integral] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle A = \int\limits^{1}_{0} {\bigg[ x \Big|\limits^{2 - y}_{y^3} \bigg]} \, dy[/tex]
- [Inner Integral] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle A = \int\limits^{1}_{0} {\big( 2 - y - y^3 \big)} \, dy[/tex]
- [Integral] Rewrite [Integration Property - Addition/Subtraction]: [tex]\displaystyle A = \int\limits^{1}_{0} {2} \, dy - \int\limits^{1}_{0} {y} \, dy - \int\limits^{1}_{0} {y^3} \, dy[/tex]
- [1st Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle A = 2\int\limits^{1}_{0} {} \, dy - \int\limits^{1}_{0} {y} \, dy - \int\limits^{1}_{0} {y^3} \, dy[/tex]
- [Integrals] Integrate [Integration Rule - Reverse Power Rule]: [tex]\displaystyle A = 2y \Big| \limits^{1}_{0} - \frac{y^2}{2} \bigg| \limits^{1}_{0} - \frac{y^4}{4} \bigg| \limits^{1}_{0}[/tex]
- [Integrals] Evaluate [Integration Rule - FTC 1]: [tex]\displaystyle A = 2 - \frac{1}{2} - \frac{1}{4}[/tex]
- Simplify: [tex]\displaystyle A = \frac{5}{4}[/tex]
∴ the area of the plane region R bounded by the curves given is 1.25.
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Learn more about integrals: https://brainly.com/question/20197752
Learn more about multivariable calculus: https://brainly.com/question/12269640
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Topic: Multivariable Calculus
Unit: Double Integrals and Area
