Mathematics, 01.03.2021 06:10, fredorivera

Please help with this question and give an explanation if possible

Answers: 3

Show answers

Answers

answered: eduardavezdemel

B. -15

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition Subtraction Left to Right

Distributive Property

Equality Properties

Multiplication Property of Equality Division Property of Equality Addition Property of Equality Subtraction Property of Equality

Calculus

Integrals

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

Integration Property [Swapping Limits]: $\displaystyle \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)} \, dx$

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

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

Integration Property [Splitting Integral]: $\displaystyle \int\limits^c_a {f(x)} \, dx = \int\limits^b_a {f(x)} \, dx + \int\limits^c_b {f(x)} \, dx$

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

Step-by-step explanation:

Step 1: Define

$\displaystyle \int\limits^3_{-1} {[2g(x) + 4]} \, dx = 22$

$\displaystyle \int\limits^{-1}_{10} {g(x)} \, dx = 12$

$\displaystyle \int\limits^{10}_{3} {g(x)} \, dx = z$

Step 2: Redefine

Manipulate the given integrals.

[Integrals] Combine [Integration Property - Splitting Integral]: $\displaystyle \int\limits^{-1}_{10} {g(x)} \, dx + \int\limits^{10}_3 {g(x)} \, dx = \int\limits^3_{10} {g(x)} \, dx$ [Integrals] Rewrite: $\displaystyle \int\limits^3_{10} {g(x)} \, dx = \int\limits^{-1}_{10} {g(x)} \, dx + \int\limits^{10}_3 {g(x)} \, dx$ [Integrals] Substitute in variables: $\displaystyle \int\limits^{-1}_3 {g(x)} \, dx = 12 + z$ [Integrals] Rewrite [Integration Property - Swapping Limits]: $\displaystyle -\int\limits^3_{-1} {g(x)} \, dx = 12 + z$ [Integrals] [Division Property of equality] Isolate integral: $\displaystyle \int\limits^3_{-1} {g(x)} \, dx = -(12 + z)$ [Integrals] [Distributive Property] Distribute negative: $\displaystyle \int\limits^3_{-1} {g(x)} \, dx = -12 - z$

Step 3: Solve

[Integral] Rewrite [Integration Property - Addition]: $\displaystyle \int\limits^3_{-1} {2g(x)} \, dx + \int\limits^3_{-1} {4} \, dx = 22$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle 2\int\limits^3_{-1} {g(x)} \, dx + 4\int\limits^3_{-1} \, dx = 22$ [Integral] Substitute in integral: $\displaystyle 2(-12 - z) + 4\int\limits^3_{-1} \, dx = 22$ [Integral] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle 2(-12 - z) + 4(x) \bigg| \limits^3_{-1} = 22$ [Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle 2(-12 - z) + 4(3 - -1) = 22$ [Integral] (Parenthesis) Simplify: $\displaystyle 2(-12 - z) + 4(3 + 1) = 22$ [Integral] (Parenthesis) Add: $\displaystyle 2(-12 - z) + 4(4) = 22$ [Integral] Multiply: $\displaystyle 2(-12 - z) + 16 = 22$ [Integral] [Subtraction Property of Equality] Subtract 16 on both sides: $\displaystyle 2(-12 - z) = 6$ [Integral] [Division Property of Equality] Divide 2 on both sides: $\displaystyle -12 - z = 3$ [Integral] [Addition Property of Equality] Isolate z term: $\displaystyle -z = 15$ [Integral] [Division Property of Equality] Isolate z: $\displaystyle z = -15$ [Integral] Back-Substitute: $\displaystyle \int\limits^{10}_{3} {g(x)} \, dx = -15$

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

Unit: Integration

Book: College Calculus 10e

answered: Guest

Dang i wish i could , good luck (i’m bad at math)
Someone its due in 30 min i dont know how to do pls