Question

Please help!! Find formula for area A(x) given graph

Answer 1

please tell me if I'm right

1. 3x
2. 1/2(x)^2
3. 5x

Answer 2

i think you might need another term for the second region

Answer 3

Can you please be more specific

Answer 4

The area of the second region is like a rectangle plus a triangle.
Your answer above, only includes the triangular section

Answer 5

I don't even think the triangular region is correct; if x = 4 then the 1/2(x)^2 would be = 4, it's only 0.5...

Answer 6

or does it just want this?
\[\int\limits_{3}^{5} f(x) dx\]

Answer 7

nvm that's not an option

Answer 8

yes, i think you would need to use definite integrals

Answer 9

I know for the second one it's 1/2(5)^2 - 1/2(3)^2 for the total area, I just don't know what to do in terms of A(x) and x

Answer 10

because it wants a formula..

Answer 11

in the third region replace x by x-5, remember that x is [5, 7]

Answer 12

okay that makes sense thanks

So for 2, would it be 1/2(x-3)^2 ?

Answer 13

yeah, kinda, but you still have to add the rectangular term for the second region
(the rectangular term for the second region is like the first region, but again with the the shift x -> x-3)

Answer 14

+ 3(x-3) ?

Answer 15

Yeah that looks better,
now check your answer, by evaluating the area formulae over the regions, and also by counting the squares. Hopefully the values will match up.

Answer 16

for the first one

1. \[A(x) = 3x\]over the region [0, 3]\[ \qquad= 3x|_0^3 = 3(3) - 3(0) = 9-0=9\]

which agrees with the count of squares in this first region

Answer 17

can you check the other two regions?

Answer 18

2. \[A(x) = 3(x-3) + \frac12(x-3)^2\]over the region [3, 5]
\[\qquad = 3(x-3) + \frac12(x-3)^2\Big|_3^5\\\qquad= 3(5-3) + \frac12(5-3)^2 -(3(3-3) + \frac12(3-3)^2)\\\qquad= 3(2) + \frac12(2)^2\\\qquad= 6+2\\\qquad=\]

Answer 19

does this agree with the count of squares?