Question

\[\int\limits_{-9}^{-6} f(x)=1 \int\limits_{-9}^{-8}f(x)=7 \int\limits_{-7}^{-6}=7 what is \int\limits_{-7}^{-8} f(x)-7 dx\]?

Answer 1

Can you write it correctly or rephrase your question please?

Answer 2

He's asking for the area between -7 and -8.

Answer 3

let \(F\) be an antiderivative of \(f\) i.e. \(F(b)-F(a)=\int_a^b f\)

Answer 4

so \(\int_{-9}^{-6}f=F(-6)-F(-9)\) whereas \(\int_{-9}^{-8}f=F(-8)-F(-9)\)

Answer 5

also \(\int_{-7}^{-6}f=F(-6)-F(-7)\)

Answer 6

we're told $$\int\limits_{-9}^{-6} f=1\\\int\limits_{-9}^{-8}f=7\\\int\limits_{-7}^{-6}f=7$$

Answer 7

correct. I have already found \[\int\limits_{-8}^{-7} f(x)= -13\]

Answer 8

we want to find

\(\int\limits_{-7}^{-8} (f(x)-7)\ dx=F(-8)-F(-7)-7(-8-(-7))=F(-8)-F(-7)+7\)

Answer 9

but just negating my answer and plugging in does not get correct answer.

Answer 10

we dont know and actuel function just their integral values.

Answer 11

observe:$$F(-8)-F(-7)=(F(-8)-F(-9))-(F(-6)-F(-9))+(F(-6)-F(-7))$$

Answer 12

and we know \(F(-8)-F(-9)=7\) while \(F(-6)-F(-9)=1\) and \(F(-6)-F(-7)=7\)

Answer 13

so \(F(-8)-F(-7)=7-1+7=13\) and thus \(\int_{-7}^{-8} (f+7)=13+7=20\)

Answer 14

\[\int\limits_{-4}^{-1} = 6 \int\limits_{-4}^{-3} = 5 \int\limits_{-2}^{-1} = 3 \int\limits_{-3}^{-2} = -2 what is \int\limits_{-2}^{-3} (6f(x) - 5) dx?\]

Answer 15

I'm not going to do an almost identical problem. The logic I used above will work the same; you should also ask this in a separate question and reward the best response to this question with a medal

Answer 16

well I tried, obviously. so thanks for the help. theres your medal