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ss7ja [257]
2 years ago
14

Use the Fundamental Theorem of Calculus to find the "area under curve" of

Mathematics
1 answer:
lozanna [386]2 years ago
7 0

Answer:

\displaystyle A = 300

General Formulas and Concepts:

<u>Calculus</u>

Integrals

  • Definite Integrals
  • Area under the curve
  • Integration Constant C

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 [Multiplied Constant]:                                                             \displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx

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

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>

<em>Identify</em>

f(x) = 6x + 19

Interval [12, 15]

<u>Step 2: Find Area</u>

  1. Substitute in variables [Area of a Region Formula]:                                       \displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx
  2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:                   \displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx
  3. [Integrals] Rewrite [Integration Property - Multiplied Constant]:                   \displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx
  4. [Integrals] Integrate [Integration Rule - Reverse Power Rule]:                      \displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:              \displaystyle A = 6(\frac{81}{2}) + 19(3)
  6. Simplify:                                                                                                             \displaystyle A = 300

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

Unit: Integration

Book: College Calculus 10e

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