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Current time:0:00Total duration:4:02

Worked example: Merging definite integrals over adjacent intervals

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)

Video transcript

what we have here is a graph of y is equal to f of X and these numbers are the areas of these shaded regions these regions between our curve and the x axis and we're going to do in this video is do some examples of evaluating definite integrals using this information and some knowledge of definite integral properties so let's start with an example let's say we want to evaluate the definite integral going from negative 4 to negative 2 of f of X D of X plus the definite integral going from negative 2 to 0 of f of X DX pause this video and see if you can evaluate this entire expression so this first part of our expression the definite integral from negative 4 to negative 2 of f of X DX we're going from x equals negative 4 to x equals negative 2 and so this would evaluate as this area between our curve and our x axis but it would be the negative of that area because our curve is below the x-axis and we could try to estimate it based on the information they've given us but they haven't given us exactly that value but we also need to figure out this right over here and here we're going from x equals negative 2 to 0 of f of X DX so that's going to be this area so if you're looking at the sum of these two definite integrals and notice the upper bound here is the lower bound here you're really thinking about this is really going to be the same thing as this is equal to the definite integral going from x equals negative 4 all the way to x equals 0 of f of X DX and this is indeed one of our integration properties if the if our upper bound here is the same as our lower bound here and we are integrating the same thing well then you can merge these two definite integrals in this way and this is just going to be this entire area but because we are below the x axis and above our curve here it would be the negative of that area so this is going to be equal to negative 7 let's do another example let's say someone were to ask you walk walk up you on the streets a quick what here's a graph what is the the the value of the expression that I'm about to write down the definite integral going from zero to four of f of X DX plus the definite integral going from four to six of f of X DX pause this video and see if you can figure that out well once again this first part right over here going from zero to four so what would be is would be this area it would be this five right over here but then we would need to subtract this area because this area is below our acts axis and above our curve we don't know exactly what this is but luckily we need also need we need to take the sum of everything I just everything I just showed but plus plus this right over here and this we're going from four to six so it's going to be this area well once again when you look at it this way you can see that this expression is going to be equivalent to taking the definite integral all the way from zero to six 0 to 6 of f of X DX and once again even if you didn't see the graph you would know that because in both cases are getting the definite integral of f of X DX and our upper bound here is our lower bound here so once again we're able to merge the integrals and what is this going to be equal to well we have this area here which is 5 and then we have this area which is 6 that was given to us but since it's below the x-axis and above our curve 1 we evaluated as a definite integral it would evaluate as a negative 6 so this is going to be 5 plus negative 6 which is equal to negative 1 and we're done