Average Value - Graphs
$begingroup$
long method: Determine an equation for each and solve using average value formula
alternative methods?
How could you prove the average value to be C over an interval [a,b] if you are given a graph.... looking for most efficient/unique methods.
calculus integration algebra-precalculus definite-integrals
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
asked Jul 30 '14 at 12:48
confusedconfused
40821019
$endgroup$
add a comment |
$begingroup$
long method: Determine an equation for each and solve using average value formula
alternative methods?
How could you prove the average value to be C over an interval [a,b] if you are given a graph.... looking for most efficient/unique methods.
calculus integration algebra-precalculus definite-integrals
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
asked Jul 30 '14 at 12:48
confusedconfused
40821019
$endgroup$
$begingroup$
In general you would integrate the function (to find the area below the curve) and divide by the length of the interval ($b-a=6$ here) but I think what is expected here is just to use a bit of guesswork: graph A clearly has an average $>2$, whereas by computing the area of triangles you can find that the average of C and D are 2 and 4 respectively. To show that B has $>2$, just draw the triangle {(0,0),(6,0),(3,4)} : it is contained in the area of the curve and has area =2
$endgroup$
– user157321
Jul 30 '14 at 12:58
add a comment |
$begingroup$
long method: Determine an equation for each and solve using average value formula
alternative methods?
How could you prove the average value to be C over an interval [a,b] if you are given a graph.... looking for most efficient/unique methods.
calculus integration algebra-precalculus definite-integrals
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
asked Jul 30 '14 at 12:48
confusedconfused
40821019
$endgroup$
long method: Determine an equation for each and solve using average value formula
alternative methods?
How could you prove the average value to be C over an interval [a,b] if you are given a graph.... looking for most efficient/unique methods.
calculus integration algebra-precalculus definite-integrals
calculus integration algebra-precalculus definite-integrals
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
asked Jul 30 '14 at 12:48
confusedconfused
40821019
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
asked Jul 30 '14 at 12:48
confusedconfused
40821019
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
edited Jul 30 '14 at 12:57
apnorton
15.2k33796
15.2k33796
asked Jul 30 '14 at 12:48
confusedconfused
40821019
asked Jul 30 '14 at 12:48
confusedconfused
40821019
asked Jul 30 '14 at 12:48
confusedconfused
40821019
40821019
$begingroup$
In general you would integrate the function (to find the area below the curve) and divide by the length of the interval ($b-a=6$ here) but I think what is expected here is just to use a bit of guesswork: graph A clearly has an average $>2$, whereas by computing the area of triangles you can find that the average of C and D are 2 and 4 respectively. To show that B has $>2$, just draw the triangle {(0,0),(6,0),(3,4)} : it is contained in the area of the curve and has area =2
$endgroup$
– user157321
Jul 30 '14 at 12:58
add a comment |
$begingroup$
In general you would integrate the function (to find the area below the curve) and divide by the length of the interval ($b-a=6$ here) but I think what is expected here is just to use a bit of guesswork: graph A clearly has an average $>2$, whereas by computing the area of triangles you can find that the average of C and D are 2 and 4 respectively. To show that B has $>2$, just draw the triangle {(0,0),(6,0),(3,4)} : it is contained in the area of the curve and has area =2
$endgroup$
– user157321
Jul 30 '14 at 12:58
$begingroup$
In general you would integrate the function (to find the area below the curve) and divide by the length of the interval ($b-a=6$ here) but I think what is expected here is just to use a bit of guesswork: graph A clearly has an average $>2$, whereas by computing the area of triangles you can find that the average of C and D are 2 and 4 respectively. To show that B has $>2$, just draw the triangle {(0,0),(6,0),(3,4)} : it is contained in the area of the curve and has area =2
$endgroup$
– user157321
Jul 30 '14 at 12:58
$begingroup$
In general you would integrate the function (to find the area below the curve) and divide by the length of the interval ($b-a=6$ here) but I think what is expected here is just to use a bit of guesswork: graph A clearly has an average $>2$, whereas by computing the area of triangles you can find that the average of C and D are 2 and 4 respectively. To show that B has $>2$, just draw the triangle {(0,0),(6,0),(3,4)} : it is contained in the area of the curve and has area =2
$endgroup$
– user157321
Jul 30 '14 at 12:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your equation for this question is:
$$frac{1}{6}*int_0^6 h(x) = 2$$
Therefore,
$$int_0^6 h(x) = 12$$
For answers $c$ and $d$ you can use the area of a triangle,
If you try answer choice $c$,
We get that:
$$A = frac{1}{2}*4.5*6 - frac{1}{2}*1.5*(-2) = 12$$
Therefore our answer is C.
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
$endgroup$
add a comment |
$begingroup$
Graph A:
The part for $xin[0,2)$ compensates with the part for $xin (2,4]$, giving an average of 2 over the interval $[0,4]$. But the part for $xin(4,6]$ is above the line $y=2$. So A is not good.
GraphB:
Define a new function $g$ whose graph is composed of one straight line between $(0,0)$ and $(3,4)$ and the other straight line between $(3,4)$ and $(6,0)$. Similarly as in case A, we can see average of the function $g$ is equal to 2. But the graph of $f$ is always above $g$. So B is not good.
Similarly, we find average in graph C is 2 and average in Graph D is 4
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your equation for this question is:
$$frac{1}{6}*int_0^6 h(x) = 2$$
Therefore,
$$int_0^6 h(x) = 12$$
For answers $c$ and $d$ you can use the area of a triangle,
If you try answer choice $c$,
We get that:
$$A = frac{1}{2}*4.5*6 - frac{1}{2}*1.5*(-2) = 12$$
Therefore our answer is C.
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
$endgroup$
add a comment |
$begingroup$
Your equation for this question is:
$$frac{1}{6}*int_0^6 h(x) = 2$$
Therefore,
$$int_0^6 h(x) = 12$$
For answers $c$ and $d$ you can use the area of a triangle,
If you try answer choice $c$,
We get that:
$$A = frac{1}{2}*4.5*6 - frac{1}{2}*1.5*(-2) = 12$$
Therefore our answer is C.
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
$endgroup$
add a comment |
$begingroup$
Your equation for this question is:
$$frac{1}{6}*int_0^6 h(x) = 2$$
Therefore,
$$int_0^6 h(x) = 12$$
For answers $c$ and $d$ you can use the area of a triangle,
If you try answer choice $c$,
We get that:
$$A = frac{1}{2}*4.5*6 - frac{1}{2}*1.5*(-2) = 12$$
Therefore our answer is C.
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
$endgroup$
Your equation for this question is:
$$frac{1}{6}*int_0^6 h(x) = 2$$
Therefore,
$$int_0^6 h(x) = 12$$
For answers $c$ and $d$ you can use the area of a triangle,
If you try answer choice $c$,
We get that:
$$A = frac{1}{2}*4.5*6 - frac{1}{2}*1.5*(-2) = 12$$
Therefore our answer is C.
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
answered Jul 30 '14 at 12:55
Varun IyerVarun Iyer
5,317826
5,317826
add a comment |
add a comment |
$begingroup$
Graph A:
The part for $xin[0,2)$ compensates with the part for $xin (2,4]$, giving an average of 2 over the interval $[0,4]$. But the part for $xin(4,6]$ is above the line $y=2$. So A is not good.
GraphB:
Define a new function $g$ whose graph is composed of one straight line between $(0,0)$ and $(3,4)$ and the other straight line between $(3,4)$ and $(6,0)$. Similarly as in case A, we can see average of the function $g$ is equal to 2. But the graph of $f$ is always above $g$. So B is not good.
Similarly, we find average in graph C is 2 and average in Graph D is 4
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
add a comment |
$begingroup$
Graph A:
The part for $xin[0,2)$ compensates with the part for $xin (2,4]$, giving an average of 2 over the interval $[0,4]$. But the part for $xin(4,6]$ is above the line $y=2$. So A is not good.
GraphB:
Define a new function $g$ whose graph is composed of one straight line between $(0,0)$ and $(3,4)$ and the other straight line between $(3,4)$ and $(6,0)$. Similarly as in case A, we can see average of the function $g$ is equal to 2. But the graph of $f$ is always above $g$. So B is not good.
Similarly, we find average in graph C is 2 and average in Graph D is 4
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
add a comment |
$begingroup$
Graph A:
The part for $xin[0,2)$ compensates with the part for $xin (2,4]$, giving an average of 2 over the interval $[0,4]$. But the part for $xin(4,6]$ is above the line $y=2$. So A is not good.
GraphB:
Define a new function $g$ whose graph is composed of one straight line between $(0,0)$ and $(3,4)$ and the other straight line between $(3,4)$ and $(6,0)$. Similarly as in case A, we can see average of the function $g$ is equal to 2. But the graph of $f$ is always above $g$. So B is not good.
Similarly, we find average in graph C is 2 and average in Graph D is 4
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
$endgroup$
Graph A:
The part for $xin[0,2)$ compensates with the part for $xin (2,4]$, giving an average of 2 over the interval $[0,4]$. But the part for $xin(4,6]$ is above the line $y=2$. So A is not good.
GraphB:
Define a new function $g$ whose graph is composed of one straight line between $(0,0)$ and $(3,4)$ and the other straight line between $(3,4)$ and $(6,0)$. Similarly as in case A, we can see average of the function $g$ is equal to 2. But the graph of $f$ is always above $g$. So B is not good.
Similarly, we find average in graph C is 2 and average in Graph D is 4
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
answered Jul 30 '14 at 13:01
Petite EtincellePetite Etincelle
12.4k12148
12.4k12148
add a comment |
add a comment |
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$begingroup$
In general you would integrate the function (to find the area below the curve) and divide by the length of the interval ($b-a=6$ here) but I think what is expected here is just to use a bit of guesswork: graph A clearly has an average $>2$, whereas by computing the area of triangles you can find that the average of C and D are 2 and 4 respectively. To show that B has $>2$, just draw the triangle {(0,0),(6,0),(3,4)} : it is contained in the area of the curve and has area =2
$endgroup$
– user157321
Jul 30 '14 at 12:58