Question

Find the area (in square units) of the region under the graph of the function f on the interval [−1, 3].

f(x) = 2x + 4

Answer 1

`\displaystyle \int\limits^3_(-1) {2x + 4} \, dx = 24`

General Formulas and Concepts:

Calculus

Integration

• 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

`\displaystyle f(x) = 2x + 4 \\\left[ -1 ,\ 3]`

Step 2: Integrate

1. Substitute in variables [Area of a Region Formula]:
   `\displaystyle \int\limits^3_(-1) {2x + 4} \, dx`
2. [Integral] Rewrite [Integration Property - Addition/Subtraction]:
   `\displaystyle \int\limits^3_(-1) {2x + 4} \, dx = \int\limits^3_(-1) {2x} \, dx + \int\limits^3_(-1) {4} \, dx`
3. [Integrals] Rewrite [Integration Property - Multiplied Constants]:
   `\displaystyle \int\limits^3_(-1) {2x + 4} \, dx = 2 \int\limits^3_(-1) {x} \, dx + 4 \int\limits^3_(-1) {} \, dx`
4. [Integrals] Integration Rule [Reverse Power Rule]:
   `\displaystyle \int\limits^3_(-1) {2x + 4} \, dx = 2 \bigg( (x^2)/(2) \bigg) \bigg| \limits^3_(-1) + 4(x) \bigg| \limits^3_(-1)`
5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^3_(-1) {2x + 4} \, dx = 2(4) + 4(4)`
6. Simplify:
   `\displaystyle \int\limits^3_(-1) {2x + 4} \, dx = 24`

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

Unit: Integration