Think about this interval right over here. Worked when both of them were above the x-axis, but what about the case when f of x is above the x-axis and g of x is below the x-axis? So for example, let's say that we were to So I know what you're thinking, you're like okay well that And now I'll make a claim to you, and we'll build a littleīit more intuition for this as we go through this video, but over an integral from a to b where f of x is greater than g of x, like this interval right over here, this is always going to be the case, that the area between the curves is going to be the integral for the x-interval that weĬare about, from a to b, of f of x minus g of x. Integration properties that we can rewrite this as the integral from a to b of, let me put some parentheses here, of f of x minus g of x, minus g of x dx. I would net out with thisĪrea right over here. With the original area that I cared about. And then if I were to subtract from that this area right over here, which is equal to that's the definite integral from a to b of g of x dx. Try to calculate this? Well one natural thing that you might say is well look, if I were to take the integral from a to b of f of x dx, that would give me the entire area below f of x and above the x-axis. So based on what you already know about definite integrals, how would you actually So that would be this area right over here. So let's say we care about the region from x equals a to x equals b between y equals f of xĪnd y is equal to g of x. We are now going to then extend this to think about the area between curves. We have already covered the notion of area betweenĪ curve and the x-axis using a definite integral. The +C represents ANY constant, not a particular constant, so it will absorb all of the constants generated by the integration.Īs mentioned above, once you become proficient at it, there is no need to show this much detail. Note: notice that I did not use two +C terms. I will include some extra steps to make things perfectly clear, but you don't actually have to show this much detail: So, here is how to solve the example you mentioned. If it is not present, you cannot use the standard integral form at all. The du of the standard integral forms is the derivative of whatever you decided to call u. You will often see beginners (and sometimes people who have enough experience to know better) ignore the du as if it were just some trivial bit of notation: it is NOT. The important thing to remember in integration is that you have to have an EXACT match of a standard integral form in order to use it. This can be a tedious process that is easy to make mistakes on, but sometimes it is what you need to do. Method 2: If it is not possible to convert the problem to a ∫u^(n) du nor to some other standard integral from, then you can expand out the polynomial and integrate each term separately. Method 1: If it is possible to convert the problem to a ∫u^(n) du form, then you can simply use the power rule. But, in general here are your best options: The exact details of the problem matter, so there cannot be a one-size-fits all solution.
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