Respuesta :

Answer:

24

Step-by-step explanation:

The question is asking for the net area from x=3 to x=10.

It gives you the net area from x=3 to x=5 being -18.

It gives you the net area from x=5 to x=10, being 42.

Together those intervals make up the interval we want to find the net area for.

-18+42=42-18=24

Space

Answer:

[tex]\displaystyle \int\limits^{10}_3 {f(t)} \, dt = 24[/tex]

General Formulas and Concepts:

Calculus

Integration

  • Integrals

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle \int\limits^{5}_3 {f(t)} \, dt = -18[/tex]

[tex]\displaystyle \int\limits^{10}_5 {f(t)} \, dt = 42[/tex]

[tex]\displaystyle \int\limits^{10}_3 {f(t)} \, dt[/tex]

Step 2: Integrate

  1. [Integral] Rewrite [Integration Property - Splitting Integral]:                      [tex]\displaystyle \int\limits^{10}_3 {f(t)} \, dt = 24 = \int\limits^5_3 {f(t)} \, dt + \int\limits^{10}_5 {f(t)} \, dt[/tex]
  2. [Integrals] Substitute:                                                                                    [tex]\displaystyle \int\limits^{10}_3 {f(t)} \, dt = 24 = -18 + 42[/tex]
  3. Simplify:                                                                                                         [tex]\displaystyle \int\limits^{10}_3 {f(t)} \, dt = 24[/tex]

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

Unit: Integration

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