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

Question

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Space Space · Answer 1 · 2021-09-10T05:45:38+02:00

Answer:

[tex]\displaystyle \int\limits^3_{-1} {2x + 4} \, dx = 24[/tex]

General Formulas and Concepts:

Calculus

Integration

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

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

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

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

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

Step-by-step explanation:

Step 1: Define

Identify

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

Step 2: Integrate

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

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

Unit: Integration