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# Definite integrals on adjacent intervals

AP.CALC:
FUN‑6 (EU)
,
FUN‑6.A (LO)
,
FUN‑6.A.1 (EK)
,
FUN‑6.A.2 (EK)
By subdividing the stretch of numbers where you are integrating, you can break up an integral.

## Want to join the conversation?

• why does c have to be in between a and b
if c was greater than b, the integral from c to b would be negative, and thus the integral from a to b would be equal to the integral from a to c minus the integral from b to c
• In later problems you will see that instead of "a" and "b" you might have "a" = to "x^2" and "b" = to the "square root of x". This makes it necessary to split it up with "c" between "a" and "b" so that the integral from "x^2"(a) to "root x" (b) is equal to the integral of "c" to "root x"(b) minus the integral "c" to "x^2"(a). This makes it possible to solve the problem.
• I don't understand he says to use this property to solve when there's a discontinuity in a function, did we not say first thing in fundamental theorem of calculus that f has to be continuous between a and b?
• You can separate the integral into two parts. So integrate from a to b, for let say for f(x). Then integrate from b to c for g(x)
• point c is calculated twice, so how will this affect (or not affect) the final outcome?