Question

numbers 7 and 8?

Answer 1

#8 you want method or answer?

Answer 2

method please

Answer 3

for most of my questions I ask them because I don't know or remember the method. just fyi

Answer 4

ok two ways. simplest way is to graph
\[y=|x-3|\] on the interval from 2 to 4
you will see two triangles so the area will be simple geometry, one half base times height

Answer 5

here is a picture, and you can see that both triangles have base 1 and height 1 so the area of each is 1/2 and the total area is 1
http://www.wolframalpha.com/input/?i=y%3D |x-3|++from+2+to+4

Answer 6

got it

Answer 7

more annoying method is to say that
\[f(x) = |x-3| = \left\{\begin{array}{rcc}
x - 3 & \text{if} & x \geq 3 \\
- x +3& \text{if} & x < 3
\end{array}
\right.\] and break the integral up into two pieces
\[\int_2^3( -x+3) dx+\int_3^4(x-3)dx\]
you can see why i prefer method one

Answer 8

#7 antiderivative is
\[\frac{1}{2}x^4-\frac{k}{3}x^3+2kx\] plug in 2 for x, set the answer = 12 and solve for k

Answer 9

you can safely ignore the lower limit because if you replace x by 0 you get 0

Answer 10

ok. thank you!

Answer 11

yw