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## Integral Calculus

### Course: Integral Calculus > Unit 1

Lesson 8: Properties of definite integrals- Negative definite integrals
- Finding definite integrals using area formulas
- Finding definite integrals using area formulas
- Definite integral over a single point
- Integrating scaled version of function
- Switching bounds of definite integral
- Integrating sums of functions
- Worked examples: Finding definite integrals using algebraic properties
- Finding definite integrals using algebraic properties
- Definite integrals on adjacent intervals
- Worked example: Breaking up the integral's interval
- Worked example: Merging definite integrals over adjacent intervals
- Definite integrals over adjacent intervals
- Functions defined by integrals: switched interval
- Finding derivative with fundamental theorem of calculus: x is on lower bound
- Finding derivative with fundamental theorem of calculus: x is on both bounds
- Functions defined by integrals: challenge problem
- Definite integrals properties review

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# Functions defined by integrals: switched interval

AP.CALC:

FUN‑6 (EU)

, FUN‑6.A (LO)

, FUN‑6.A.1 (EK)

, FUN‑6.A.2 (EK)

Sal evaluates a function defined by the integral of a graphed function. In order to evaluate he must switch the sides of the interval.

## Want to join the conversation?

- Why did Sal put a negative next to the integral sign?(10 votes)
- I remember this by thinking of a definite integral as just subtracting two numbers. If we assume a wholly positive function for simplicity, think of the definite integral as the area from negative infinity to the top terminal minus the area from negative infinity to the bottom terminal. If the top number is bigger than the bottom number, the negative-infinity-to-top-terminal area is going to be bigger than the negative-infinity-to-bottom-terminal area, and for a positive function you get a positive result, no problem. However, when the bottom number is larger than the top number, the negative-infinity-to-top-terminal area is smaller than the negative-infinity-to-bottom-terminal area. If you think of the areas as just numbers, you realise you are subtracting a larger number from a smaller number and you are going to get a negative answer. Just like the when comparing 5-3 = 2 and 3-5 = -2, the "distance" between the numbers is the same, only one of the answers is negative.(7 votes)

- why is it negative? it is above the x axis.(4 votes)
- Sal did not make a mistake here. Recall that when switching bounds of a definite integral, the answer will be the negative form of the original value.(12 votes)

- In the excersises for def int reverse power rule, there are a couple of problems where the bounds are reversed (such as a=2, b=1 12x^-5). But we're not multiplying the solution by -1 to account for the switched intervals. Am i missing something (only seems to happen when the exonent is negative)(3 votes)
- what if it asked you for the equation of the line at that point? How would you find the slope?(2 votes)
- What is the difference between switching intervals and switching bounds?(2 votes)
- No difference in this context.(2 votes)

## Video transcript

- [Voiceover] The graph
of f is shown below. Let g(x) be equal to the def intergral from zero to x, of f(t)dt. Now at first when you
see this you're like, wow this is strange, I have a function that is being defined by an integral, a def integral, but one
of it's bounds are X. You should just say well this is okay, a function can be defined any which way, and as we'll see, it's
actually quite straight forward to evaluate this. So g of negative two, g of negative two, and I'll do the negative
two in a different color, g of negative two, what we do is take this
expression over here, this def integral, and
where ever we see an X, we replace it with a negative two. So this is going to be equal to, the integral from zero to X, and I'll write X in a second, f(t)dt. Well X is now negative two,
this is now negative two. So how do we figure out what this is? Before we even look at this graph, you might say okay this
is the region under, the area of the region under the graph, y equals f(t), between
negative two and zero. But you have to be careful, notice, our upper bound here is
actually a lower number than our lower bound right over here. So it will be nice to swap those bounds so we can truly view it
as the area of the region under f(t), above the t axis,
between those two bounds. When you swap the bounds,
this is going to be equal to negative def integral, from negative two, negative two to zero, of f(t)dt, and now what we have right over here, what I'm squaring off in magenta, this is the area under the curve f, between negative two and zero. So between negative two and zero, so that is this area, right over here, that we care about. Now what is that going to be? There's a bunch of different
ways that you could do this, you could split it off
into a square and triangle. The area of this square here is four, it's two by two, make
sure to look at the units, sometimes each square doesn't
represent one square unit, in this case it does, so that's four, and then up here, this is half of four. If it was all of this, that would be four, this triangle is half
of four, so this is two. Or you could view this as base
times height times one half, which is going to be two
times two times one half. So this area is six, so this part is six, but we can't forget that negative sign, so this is going to be
equal to negative six. So g(-2) is negative six.