- Integrating scaled version of function
- Integrating sums of functions
- Definite integral over a single point
- Definite integrals on adjacent intervals
- Definite integral of shifted function
- Switching bounds of definite integral
- Worked examples: Finding definite integrals using algebraic properties
- Using multiple properties of definite integrals
- Definite integral properties (no graph): function combination
- Worked examples: Definite integral properties 2
- Definite integral properties (no graph): breaking interval
- Warmup: Definite integral properties (no graph)
- Finding definite integrals using algebraic properties
- Examples leveraging integration properties
- Definite integrals properties review
Given the definite integrals of f and g over a specific interval, Sal finds the definite integral (on that interval) of a combination of f and g.
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- Thank you very much for your video. At the 0.45, do I need put closing bracket before
- 𝑑𝑥 is treated pretty much like any other algebraic quantity, which means that if we have an expression with multiple terms and want to multiply it by 𝑑𝑥 we put brackets around the expression.
∫(𝑥𝑒ˣ)𝑑𝑥 = ∫𝑥𝑒ˣ𝑑𝑥
∫(𝑥 + 𝑒ˣ)𝑑𝑥 ≠ ∫𝑥 + 𝑒ˣ𝑑𝑥(3 votes)
- what is the difference between definite integral and indefinite integral?(1 vote)
- indefinite integral is boundless (I don't know if that's the right word) it is not well defined. for ex integral of sinx is -cosx + c and why does this c matter? bc as we know integration is basically area under the curve (an easy way to approach) so when the value of c changes, the area under the curve changes as well. this is where definite integral comes to play, here there's no c as it is already bounded to some limit. so there's certainty in the answer obtained, unlike indefinite where we don't know what c might be, which is uncertainty.
this is how I like to approach, please do correct me if i'm not :)(1 vote)
- [Voiceover] Given that the definite integral from negative one to three of f of x dx is equal to negative two and the definite integral from negative on to three of g of x dx is equal to five, what is the definite integral from negative one to three of three f of x minus two g of x dx? All right, so to think about this what we could use is some of our integration properties. And so the first thing that I would want to do is we could split this up into two integrals. We know that the and this is true of definite or indefinite integrals, that the integral of f of x the integral of f of x plus or minus g of x dx is going to be equal to the integral of f of x dx plus or minus the integral of g of x dx. If this is a plus, this is gonna be a plus. If this is a minus, this is going to be a minus. So we could split this up in the same way. So this is going to be equal to the definite integral from negative one to three of three f of x dx minus the integral from negative one to three of two g of x dx. Notice, all I did is I split it up. Taking the integral of the difference of these functions is the same thing as taking the difference of the integrals of those functions. Now the next thing we can do is we can take the scalars, we're multiplying the functions on the inside by these numbers, three and two, and we can take those outside of the integral. And that comes straight out of the property that if I'm taking the integral of some constant times f of x dx, that is equal to the constant times the integral of f of x dx, and so I can rewrite this as, so, let's see, I could rewrite this first integral as three times the definite integral from negative one to three of f of x dx minus two times the definite integral from negative one to three of g of x, actually let me do the second one in a different color. Minus, minus, this is gonna be in magenta. Minus two times the integral from negative one to three of g of x dx. And so what is this going to be equal to? Well they tell us, they tell us what this thing is here that I'm underlining in orange, the integral from negative one to three of f of x dx, they tell us that that is equal to negative two. So that thing is negative two, and likewise, this thing right over here, the definite integral from negative one to three of g of x dx, they gave it right over here, it's equal to five. So that's equal to five, and so the whole thing is going to be three times negative two, which is equal to negative six minus two times five minus 10, which is equal to negative 16. And we're done!