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## Calculus 2

### Course: Calculus 2 > Unit 1

Lesson 8: Properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Functions defined by integrals: challenge problem
- Definite integrals properties review

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# Definite integrals properties review

Review the definite integral properties and use them to solve problems.

## What are the definite integral properties?

**Sum/Difference:**

*Want to learn more about this property? Check out this video.*

**Constant multiple:**

*Want to learn more about this property? Check out this video.*

**Reverse interval:**

*Want to learn more about this property? Check out this video.*

**Zero-length interval:**

*Want to learn more about this property? Check out this video.*

**Adding intervals:**

*Want to learn more about this property? Check out this video.*

## Practice set 1: Using the properties graphically

*Want to try more problems like this? Check out this exercise.*

## Practice set 2: Using the properties algebraically

*Want to try more problems like this? Check out this exercise.*

## Want to join the conversation?

- How would I graph an equation like g(x)=3f(x-2)+1?(2 votes)
- You need to have some info about f(x). I mean how do you graph a function that depends on another function that has not been defined.(7 votes)

- If the f(x) inside is f(3x) instead of 3f(x), do you multiply the 3 to the bound values (a and b)? (does then become F(3b)- F(3a)?)(4 votes)
- Proof of.

|f(x)g(x)d(x)=|f(c)g(x)d(x)

c€[a,b].

|~integration from a-b(1 vote)- ∫(𝑎, 𝑏) 𝑓(𝑥)𝑑𝑥 and ∫(𝑎, 𝑏) 𝑓(𝑐)𝑑𝑥 are not equivalent expressions.

Example:

𝑎 = 0, 𝑏 = 2

𝑓(𝑥) = 𝑥 ⇒ ∫(𝑎, 𝑏) 𝑓(𝑥)𝑑𝑥 = ∫(0, 2) 𝑥𝑑𝑥 = 2²∕2 − 0²∕2 = 2

𝑐 = 2𝑥 ⇒ 𝑓(𝑐) = 2𝑥 ⇒ ∫(𝑎, 𝑏) 𝑓(𝑐)𝑑𝑥 = ∫(0, 2) 2𝑥𝑑𝑥 = 2 ∙ ∫(0, 2) 𝑥𝑑𝑥 = 2 ∙ 2 = 4(6 votes)

- how do u integrate xdx between bounds x and 0(1 vote)
- It'll use the reverse power rule. You integrate xdx to x^(2)/2. Then, substitute the bounds to get 0 - x^(2)/2 = -x^(2)/2(1 vote)

- integral f(x) from 3a to 3b equal to integral 3f(3x)from a to b. is this right?(0 votes)
- No, this is not true. Integrating from 3a to 3b would mean you are changing the bounds of integration - it totally depends on what the function looks like over the interval from x=3a to x=3b. It may look the same as it does over the interval from x=a to x=b, but odds are it doesn't.(2 votes)

- The last property should be ∫(𝑎, 𝑏) 𝑓(𝑥)𝑑𝑥 = ∫(𝑎, c) 𝑓(𝑥)𝑑𝑥 + ∫(c, 𝑏) 𝑓(𝑥)𝑑𝑥(0 votes)
- Note that they're the same—here b replaces the c in your formula.(1 vote)

- $\sqrt{x}$ does this work?///

$$ \int_0^8 f(x) = \left[ x + \frac{1}{10} \cdot \frac{x^3}{3} \right]_0^8 = 12 + \frac{12^3}{30} - 0 = 12+56.6 = 69.6 $$(0 votes)