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Musya8 [376]
2 years ago
7

Area of the bounded curves y=x^2, y=√(7+x)

Mathematics
1 answer:
N76 [4]2 years ago
5 0

Answer:

\displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 5.74773

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Addition/Subtraction]:                                                         \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]  

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:                                                               \displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C

Integration Rule [Fundamental Theorem of Calculus 1]:                                     \displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)

Integration Property [Addition/Subtraction]:                                                       \displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

U-Substitution

Area of a Region Formula:                                                                                     \displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx

Step-by-step explanation:

<u>Step 1: Define</u>

\displaystyle \left \{ {{y = x^2} \atop {y = \sqrt{7 + x}}} \right.

<u>Step 2: Identify</u>

<em>Graph the systems of equations - see attachment.</em>

Top Function:  \displaystyle y = \sqrt{7 + x}

Bottom Function:  \displaystyle y = x^2

Bounds of Integration: [-1.529, 1.718]

<u>Step 3: Integrate Pt. 1</u>

  1. Substitute in variables [Area of a Region Formula]:                                   \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:               \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - \int\limits^{1.718}_{-1.529} {x^2} \, dx
  3. [Right Integral] Integration Rule [Reverse Power Rule]:                             \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - \frac{x^3}{3} \bigg| \limits^{1.718}_{-1.529}
  4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{1.718}_{-1.529} {\sqrt{7 + x}} \, dx - 2.88176

<u>Step 4: Integrate Pt. 2</u>

<em>Identify variables for u-substitution.</em>

  1. Set <em>u</em>:                                                                                                             \displaystyle u = 7 + x
  2. [<em>u</em>] Basic Power Rule [Derivative Rule - Addition/Subtraction]:                 \displaystyle du = dx
  3. [Limits] Switch:                                                                                               \displaystyle \left \{ {{x = 1.718 ,\ u = 7 + 1.718 = 8.718} \atop {x = -1.529 ,\ u = 7 - 1.529 = 5.471}} \right.

<u>Step 5: Integrate Pt. 3</u>

  1. [Integral] U-Substitution:                                                                               \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx= \int\limits^{8.718}_{5.471} {\sqrt{u}} \, du - 2.88176
  2. [Integral] Integration Rule [Reverse Power Rule]:                                       \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = \frac{2x^\Big{\frac{3}{2}}}{3} \bigg| \limits^{8.718}_{5.471} - 2.88176
  3. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 8.62949 - 2.88176
  4. Simplify:                                                                                                         \displaystyle \int\limits^{1.718}_{-1.529} {\sqrt{7 + x} - x^2} \, dx = 5.74773

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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