Question

Please help with this question and give an explanation if possible

Answer 1

B. -15

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

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 = (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)`

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.

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

Step 3: Solve

1. [Integral] Rewrite [Integration Property - Addition]:
   `\displaystyle \int\limits^3_(-1) {2g(x)} \, dx + \int\limits^3_(-1) {4} \, dx = 22`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle 2\int\limits^3_(-1) {g(x)} \, dx + 4\int\limits^3_(-1) \, dx = 22`
3. [Integral] Substitute in integral:
   `\displaystyle 2(-12 - z) + 4\int\limits^3_(-1) \, dx = 22`
4. [Integral] Integrate [Integration Rule - Reverse Power Rule]:
   `\displaystyle 2(-12 - z) + 4(x) \bigg| \limits^3_(-1) = 22`
5. [Integral] Evaluate [Integration Rule - FTC 1]:
   `\displaystyle 2(-12 - z) + 4(3 - -1) = 22`
6. [Integral] (Parenthesis) Simplify:
   `\displaystyle 2(-12 - z) + 4(3 + 1) = 22`
7. [Integral] (Parenthesis) Add:
   `\displaystyle 2(-12 - z) + 4(4) = 22`
8. [Integral] Multiply:
   `\displaystyle 2(-12 - z) + 16 = 22`
9. [Integral] [Subtraction Property of Equality] Subtract 16 on both sides:
   `\displaystyle 2(-12 - z) = 6`
10. [Integral] [Division Property of Equality] Divide 2 on both sides:
    `\displaystyle -12 - z = 3`
11. [Integral] [Addition Property of Equality] Isolate z term:
    `\displaystyle -z = 15`
12. [Integral] [Division Property of Equality] Isolate z:
    `\displaystyle z = -15`
13. [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