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AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 6
Lesson 6: Applying 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
- Definite integrals properties review
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Definite integrals properties review
AP.CALC:
FUN‑6 (EU)
, FUN‑6.A (LO)
, FUN‑6.A.1 (EK)
, FUN‑6.A.2 (EK)
Review the definite integral properties and use them to solve problems.
What are the definite integral properties?
Sum/Difference: integral, start subscript, a, end subscript, start superscript, b, end superscript, open bracket, f, left parenthesis, x, right parenthesis, plus minus, g, left parenthesis, x, right parenthesis, close bracket, d, x, equals, integral, start subscript, a, end subscript, start superscript, b, end superscript, f, left parenthesis, x, right parenthesis, d, x, plus minus, integral, start subscript, a, end subscript, start superscript, b, end superscript, g, left parenthesis, x, right parenthesis, d, x
Want to learn more about this property? Check out this video.
Constant multiple: integral, start subscript, a, end subscript, start superscript, b, end superscript, k, dot, f, left parenthesis, x, right parenthesis, d, x, equals, k, integral, start subscript, a, end subscript, start superscript, b, end superscript, f, left parenthesis, x, right parenthesis, d, x
Want to learn more about this property? Check out this video.
Reverse interval: integral, start subscript, a, end subscript, start superscript, b, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, minus, integral, start subscript, b, end subscript, start superscript, a, end superscript, f, left parenthesis, x, right parenthesis, d, x
Want to learn more about this property? Check out this video.
Zero-length interval: integral, start subscript, a, end subscript, start superscript, a, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, 0
Want to learn more about this property? Check out this video.
Adding intervals: integral, start subscript, a, end subscript, start superscript, b, end superscript, f, left parenthesis, x, right parenthesis, d, x, plus, integral, start subscript, b, end subscript, start superscript, c, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, integral, start subscript, a, end subscript, start superscript, c, end superscript, f, left parenthesis, x, right parenthesis, d, x
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?(3 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)
- $\sqrt{x}$ 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)