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Definite integral of rational function

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.B (LO)
,
FUN‑6.C (LO)
Sal finds the definite integral of (16-x³)/x³ between -1 and -2 using the reverse power rule.

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• Shouldn't it be - 7, as the interval for applying fund.th.of calculus should be [-2; -1] and not [-1; -2]?
• If you flip the bounds and add a negative sign to the entire thing (because you flipped the bounds), you should still get 7. The calculation is below:

- [[-8(-1)^(-2) - (-1)] - [-8(-2)^(-2) - (-2)]]
= - [[-7] - [0]]
= - [-7]
= 7

Intuitively, we are moving backwards from -1 to -2 (so the integral should be negative) but we are also below the x-axis (so the integral should be negative again). The two negatives cancel out and we are left with a positive answer.
• Why don't I have to account for a C constant when finding the antiderivative when it is a definite integral?
• If we have a function 𝒇(𝑥) and know its anti-derivative is 𝑭(𝑥) + C, then the definite integral from 𝑎 to 𝑏 is given by 𝑭(𝑏) + C - (𝑭(𝑎) + C). So we don't have to account for it because it cancels out.
• I learned that when we have integral from lower bound that's greater than the upper bound to that upper bound, which is less than the lower bound, we flip the bounds and have a negative integral. Why isn't that the case in this example?
• It turns out that your method is also correct, even though the way it was solved in the video was different. I also solved (and rewrote) the integral with the lower bound of -2 and an upper bound of -1. The solution is 7.
• Should we check if the function is continuous in the interval before we calculate the integration? If we calculate [-1,1], there is zero inside which is not defined, will the result be wrong?
• It doesn't matter. Point discontinuities don't affect the value of an integral, since the rectangle of area being lost has width 0.
• I know that in this video, the simple algebraic manipulation of (a+b)/b can be used, but what if we encounter say b/(a+b) and is not the antiderivative of a trig function?
• In one of the practice exercises I had to find the def. integral from -1 to 1 of [(27/x^4)-3]dx and I got the result -24 wich was the correct answer for the exercise, however, when I graphed the function on desmos.com/calculator, the area under the curve looks like positive and either very large or infinite, so what is the truth?
• I would say that -24 is wrong. And that you should let KA know that they have an incorrect question/answer, if possible.

In order for the Fundamental Theorem of Calculus to hold, the integrand has to be continuous on the interval. This is not the case for 27/𝑥⁴ - 3, which has a discontinuity at 𝑥 = 0.
• Is the notation for evaluating at bounds a official? I've never seen it before but it seems useful
• Shouldn't the answer be 7 instead of -7? When solving the problem myself, I got 7 and wasn't sure where I went wrong. So, I double checked with a graphing calculator which is also telling me that the answer is 7.
• I'm looking at the graph of (16-x^3)/x^3 and I'm confused about what this integral means on the graph. The area of the given interval lies under the x axis, not above it. This suggests that the answer should be negative, however negative areas don't exist. Or do they? I remember from trigonometry that negative angles existed, so perhaps this is a similar concept i.e doing equations with numbers that can only be expressed in equations, but not real life.

Furthermore, the answer was positive anyway, but I'm not entirely sure why. Again, I understand the equation that Sal did in the video, but I don't see how to visualize it on the graph. Is the answer positive because the upperbound was lower than the lower bound, which gives an inverse area?

Can someone please confirm that I'm understanding this correctly.