Question

Ap calculus need help please

Answer 1

`\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx = 11`

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

Algebra I

• Functions
• Function Notation

Calculus

Integrals

• Definite Integrals

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`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^1_0 {f(x)} \, dx = 4`

`\displaystyle \int\limits^1_0 {g(x)} \, dx = -3`

`\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx`

Step 2: Find

1. [Integral] Rewrite [Integration Property - Subtraction]:
   `\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx = \int\limits^1_0 {2f(x)} \, dx - \int\limits^1_0 {g(x)} \, dx`
2. [1st Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx = 2\int\limits^1_0 {f(x)} \, dx - \int\limits^1_0 {g(x)} \, dx`
3. Substitute in integral values:
   `\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx = 2(4) - (-3)`
4. Multiply:
   `\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx = 8 - (-3)`
5. Subtract:
   `\displaystyle \int\limits^1_0 {[2f(x) - g(x)]} \, dx = 11`

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

Unit: Integration

Book: College Calculus 10e