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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?
• 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.
• why is it negative? it is above the x axis.
• 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.