Question

Please show work it’s for calc

Answer 1

24

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

Answer 2

`\displaystyle \int\limits^(10)_3 {f(t)} \, dt = 24`

General Formulas and Concepts:

Calculus

Integration

• Integrals

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`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^(5)_3 {f(t)} \, dt = -18`

`\displaystyle \int\limits^(10)_5 {f(t)} \, dt = 42`

`\displaystyle \int\limits^(10)_3 {f(t)} \, dt`

Step 2: Integrate

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

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

Unit: Integration