How would you use the Fundamental Theorem of Calculus to determine the value(s) of b if the area under the graph g(x)=4x between x=1 and x=b is equal to 240?

Question

asked Mar 6, 2018 31.6k views

1 Answer

Christian Westman · Answer 1 · 2018-03-11T00:51:02+0000

Answer:

$\displaystyle b = 11$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Equality Properties

Algebra I

Functions

Calculus

Integration

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

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

Explanation:

Step 1: Define

Identify

g(x) = 4x

Interval [1, b]

A = 240

Step 2: Solve for b

Substitute in variables [Area of a Region Formula]:
$\displaystyle \int\limits^b_1 {4x} \, dx = 240$
[Integral] Rewrite [Integration Property - Multiplied Constant]:
$\displaystyle 4\int\limits^b_1 {x} \, dx = 240$
[Integral] Integrate [Integration Rule - Reverse Power Rule]:
$\displaystyle 4((x^2)/(2)) \bigg| \limits^b_1 = 240$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
$\displaystyle 4((b^2)/(2) - (1)/(2)) = 240$
[Distributive Property] Distribute 4:
$\displaystyle 2b^2 - 2 = 240$
[Addition Property of Equality] Add 2 on both sides:
$\displaystyle 2b^2 = 242$
[Division Property of Equality] Divide 2 on both sides:
$\displaystyle b^2 = 121$
[Equality Property] Square root both sides:
$\displaystyle b = \pm 11$
Choose:
$\displaystyle b = 11$

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

Unit: Integration

Book: College Calculus 10e