You do know that a definite integral gives you the area under the curve for f(x), right?
Did you draw the graph? All you have to do is figure the area of two triangles.
Use geometry(geometric formulas) to evaluate the integral from 0 to 6 f(x) dx, for the system of equations:
f(x)= 4-x for x LESS than or EQUAL to 4 and
equals 2x-8 for x > 4.
A. 24
B.8
C.16
D.12
I could really use some help please!
Thank you!
8 answers
Did you make your sketch?
Mine has two right-angled triangles.
1. for f(x) = 4-x from 0 to 4 --- triangle with vertices (0,0), (4,0) and (0,4)
2. for f(x) = 2x - 8 for x>4 ---- triangle with vertices (4,0), (6,0), and (6,4)
Area = (1/2)(4)(4) + (1/2)(2)(4) = 12
If we want to use "overkill" and apply Calculus , we would get
∫ (4-x) dx from 0 to 4 + ∫ (2x-8) dx from 4 to 6
= 4x - x^/2| from 0 to 4 + x^2 - 8x | from 4 to 6
= (16 - 8 - (0-0) ) + (36 - 48 - (16 - 32) )
= 8 + 4
= 12
Mine has two right-angled triangles.
1. for f(x) = 4-x from 0 to 4 --- triangle with vertices (0,0), (4,0) and (0,4)
2. for f(x) = 2x - 8 for x>4 ---- triangle with vertices (4,0), (6,0), and (6,4)
Area = (1/2)(4)(4) + (1/2)(2)(4) = 12
If we want to use "overkill" and apply Calculus , we would get
∫ (4-x) dx from 0 to 4 + ∫ (2x-8) dx from 4 to 6
= 4x - x^/2| from 0 to 4 + x^2 - 8x | from 4 to 6
= (16 - 8 - (0-0) ) + (36 - 48 - (16 - 32) )
= 8 + 4
= 12
Yes, I know that the definite integral gives me the area under the curve for f(x). BUT...... my only problem is that I don't understand how to do this problem at all because in my module/chapter, I've never ran in to a problem like this using a system of equations. This is legit the first time they've given me this kind of problem.
@Reiny do you mind helping me out on one more question please?
Thank you.
Thank you.
go for it
Okay I posted it. Thank you sooo much man! I really do appreciate it!
Similar to Reiny:
Draw a Cartesian coordinate system x = 0 to 6
Draw line y = 4 - x , where x = 0 to 4
For x = 0 , y = 4 , for x = 4 y = 0
Draw line y = 2 x - 8 , where x = 4 to 6
For x = 4 , y = 0 , for x = 6, y = 4
The definite integral is area of the region bounded by the graph and x-axis.
In this case:
A1 = one half area of square ( 4 * 4 ) betwen x = 0 and x = 4
and
A2 = one half area of rectangle ( 2 * 4 ) betwen x = 4 and x = 6
A = A1 + A2
A = 4 ∙ 4 / 2 + 2 ∙ 4 / 2
A = 16 / 2 + 8 / 2
A = 8 + 4 = 12
Answer 12
__________________________________________
Remark:
You have A2 = one half area of rectangle ( 2 * 4 ) becouse:
one side of rectangle:
x = 6 - 4 = 2
other side of rectanlge:
y = 4 - 0 = 4
___________________________________________
Draw a Cartesian coordinate system x = 0 to 6
Draw line y = 4 - x , where x = 0 to 4
For x = 0 , y = 4 , for x = 4 y = 0
Draw line y = 2 x - 8 , where x = 4 to 6
For x = 4 , y = 0 , for x = 6, y = 4
The definite integral is area of the region bounded by the graph and x-axis.
In this case:
A1 = one half area of square ( 4 * 4 ) betwen x = 0 and x = 4
and
A2 = one half area of rectangle ( 2 * 4 ) betwen x = 4 and x = 6
A = A1 + A2
A = 4 ∙ 4 / 2 + 2 ∙ 4 / 2
A = 16 / 2 + 8 / 2
A = 8 + 4 = 12
Answer 12
__________________________________________
Remark:
You have A2 = one half area of rectangle ( 2 * 4 ) becouse:
one side of rectangle:
x = 6 - 4 = 2
other side of rectanlge:
y = 4 - 0 = 4
___________________________________________
This is not a "system" of equations.
Remember back in pre-cal, where you studied piece-wise functions?
Remember back in pre-cal, where you studied piece-wise functions?
2 answers