Question

A curve is represented by the parametric equations x=t , y=3t^2 find the area under the curve from t=1 to t=2

Answer 1

Answer:

$\displaystyle A = 7$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Pre-Calculus

• Parametric to Rectangular Form Conversion

Calculus

Integrals - Area under the curve

Area of a Curve Formula:                                                                                     $\displaystyle A = \int\limits^b_a {f(x)} \, dx$

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

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

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

Step-by-step explanation:

*Note:

The area under the curve is essentially the definition of an integral.

Step 1: Define

Parametric

x = t

y = 3t²

1 ≤ t ≤ 2

Step 2: Rewrite

Rewrite parametric to rectangular form and change bounds of integration.

1. [Parametric] Substitute in t:                                                                            y = 3x²
2. [Parametric] Plug in values of t [Bounds]:                                                        1 ≤ x ≤ 2

Step 3: Find Area

Integration.

1. Substitute in variables/function [Area]:                                                       $\displaystyle A = \int\limits^2_1 {3x^2} \, dx$
2. [Area] Rewrite [Integration Property - Multiplied Constant]:                       $\displaystyle A = 3\int\limits^2_1 {x^2} \, dx$
3. [Area] Integrate [Integration Rule - Reverse Power Rule]:                         $\displaystyle A = 3(\frac{x^3}{3}) \bigg| \limit^2_1$
4. [Area] Evaluate [Integration Rule - Fundamental Theory of Calculus 1]:   $\displaystyle A = 3(\frac{2^3}{3} - \frac{1^3}{3})$
5. [Area] (Parenthesis) [Fraction] Evaluate exponents:                                   $\displaystyle A = 3(\frac{8}{3} - \frac{1}{3})$
6. [Area] (Parenthesis) Subtract:                                                                       $\displaystyle A = 3(\frac{7}{3})$
7. [Area] Multiply:                                                                                               $\displaystyle A = 7$

Topic: AP Calculus AB/BC

Unit: Area under the curve

Book: College Calculus 10e