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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: ${\int }_{a}^{b}\left[f\left(x\right)±g\left(x\right)\right]dx={\int }_{a}^{b}f\left(x\right)dx±{\int }_{a}^{b}g\left(x\right)dx$
Constant multiple: ${\int }_{a}^{b}k\cdot f\left(x\right)dx=k{\int }_{a}^{b}f\left(x\right)dx$
Reverse interval: ${\int }_{a}^{b}f\left(x\right)dx=-{\int }_{b}^{a}f\left(x\right)dx$
Zero-length interval: ${\int }_{a}^{a}f\left(x\right)dx=0$
Adding intervals: ${\int }_{a}^{b}f\left(x\right)dx+{\int }_{b}^{c}f\left(x\right)dx={\int }_{a}^{c}f\left(x\right)dx$

## Practice set 1: Using the properties graphically

Problem 1.1
${\int }_{-2}^{0}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx+{\int }_{0}^{3}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=$
units${}^{2}$

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

## Practice set 2: Using the properties algebraically

Problem 2.1
${\int }_{-1}^{3}f\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=-2$
${\int }_{-1}^{3}g\left(x\right)\phantom{\rule{0.167em}{0ex}}dx=5$
${\int }_{-1}^{3}\left(3f\left(x\right)-2g\left(x\right)\right)\phantom{\rule{0.167em}{0ex}}dx\phantom{\rule{0.167em}{0ex}}=$

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?
• 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.
• 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)?)
• 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
• 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?
• $\sqrt{x}$ does this work?///
$$\int_0^8 f(x) = \left[ x + \frac{1}{10} \cdot \frac{x^3}{3} \right]_0^8 = 12 + \frac{12^3}{30} - 0 = 12+56.6 = 69.6$$