Find the area of the region between the curve x^3+2x^2-3x and the x-axis over the interval [-3,1]

Question

asked Jan 18, 2022 184k views

1 Answer

InnoSPG · Answer 1 · 2022-01-24T22:01:24+0000

Answer:

$\displaystyle A = (32)/(3)$

General Formulas and Concepts:

Calculus

Integrals

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

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:
$\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Explanation:

Step 1: Define

Identify

Curve: x³ + 2x² - 3x

Interval: [-3, 1]

Step 2: Find Area

Set up:
$\displaystyle A = \int\limits^1_(-3) {(x^3 + 2x^2 - 3x)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]:
$\displaystyle A = \int\limits^1_(-3) {x^3} \, dx + \int\limits^1_(-3) {2x^2} \, dx - \int\limits^1_(-3) {3x} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]:
$\displaystyle A = \int\limits^1_(-3) {x^3} \, dx + 2\int\limits^1_(-3) {x^2} \, dx - 3\int\limits^1_(-3) {x} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]:
$\displaystyle A = ((x^4)/(4)) \bigg| \limits^1_(-3) + 2((x^3)/(3)) \bigg| \limits^1_(-3) - 3((x^2)/(2)) \bigg| \limits^1_(-3)$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
$\displaystyle A = -20 + 2((28)/(3)) - 3(-4)$
Evaluate:
$\displaystyle A = (32)/(3)$

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

Unit: Integration

Book: College Calculus 10e