1. ## Cal I Problem...

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.

2. ## Re: Cal I Problem...

Originally Posted by 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.
The idea is that f(x) has an anti-derivative F(x) such that

$\displaystyle \int_a^b f(x)dx=F(b)-F(a)$

from the first equation $\displaystyle F(2)-F(0)=1$

and from the second $\displaystyle F(6)-F(0)=6$

finally

$\displaystyle \int_6^2 f(x)dx=F(2)-F(6)=F(2)-F(6)-F(0)+F(0)=$

$\displaystyle \left(F(2)-F(0)\right)-\left(F(6)-F(0)\right)=1-6=-5$

3. ## Re: Cal I Problem...

Originally Posted by JenniferC
Given that (image attached)
What is the value of (image attached)
Here is another way:
$\displaystyle \int_0^6 {f(x)dx} = \int_0^2 {f(x)dx + } \int_2^6 {f(x)dx} = \int_0^2 {f(x)dx }\color{blue} - \int_6^2 {f(x)dx}$