Cal I Problem

[JenniferC]

Given that (image attached)

What is the value of (image attached)



Drawing the line and plugging in the point distances, I thought the answer would be 6, but that doesn't seem to be an answer choice. I'd appreciate help with this. I'd be grateful for a step-by-step guide.


Edit: Nevermind, I figured it out.

 

[JeffM]

Given that (image attached)
View attachment 3710
What is the value of (image attached)



Drawing the line and plugging in the point distances, I thought the answer would be 6, but that doesn't seem to be an answer choice. I'd appreciate help with this. I'd be grateful for a step-by-step guide.

\(\displaystyle \displaystyle \int f(x)\ dx = F(x) + C.\)

\(\displaystyle \displaystyle \int_p^q f(x)\ dx = F(q) + C - \{F(p) + C\} = F(q) - F(p).\) With me to there?

\(\displaystyle So\ \displaystyle \int_q^p f(x)\ dx = F(p) - F(q) = -\{F(q) - F(p)\} = - \int_p^q f(x)\ dx.\) Right?

OK Now watch this little manipulation.

\(\displaystyle F(r) - F(p) = F(r) - F(q) + F(q) - F(p) = \{F(q) - F(p)\} + \{F(r) - F(q)\}\implies \displaystyle \int_p^r f(x)\ dx = \int_p^q f(x)\ dx + \int_q^r f(x)\ dx.\)

And this one

\(\displaystyle \displaystyle \int_p^r f(x)\ dx = \int_p^q f(x)\ dx + \int_q^r f(x)\ dx \implies \int_p^r f(x)\ dx - \int_p^q f(x)\ dx = \int_q^r f(x)\ dx.\)

Now with these general rules, try to solve your problem again. What do you get? Is it one of the suggested answers?