Q - Combined Area?? (1 Viewer)

[Smile12345]

Hello All

If I have the line y = 5x + 4 and the parabola y = (x - 4)^2

How do I find the area between the x axis (on the bottom), the y axis (on the LH side), y = 5x + 4 (on the top) and the parabola on the RH side?? Hope this makes sense!

Thanks in advance.

 

[Squar3root]

sorry it doesn't make sense, can you say the limits (or is that not given in the question)

 

[Smile12345]

sorry it doesn't make sense, can you say the limits (or is that not given in the question)

No they're not given.... If you draw the two equations given... It's not inside the parabola... It's to the left of the parabola... To the y axis and the bottom is the x axis... Can you picture it??

 

[Study notes]

So the area bounded by the two graphs, the x and y axis?

If so...

Find the point of intersection if u solve 5x + 4 = (x - 4)^2

U take the smaller x coordinate, since u can visually see that it would make sense if u graph it.

I got x = 1.

Now if u graphed it, draw a line down the point of intersection.

Find the Integral between 0 and 1 of the line y = 5x + 4

then find the integral between 1 and 4 (since that's where the parabola touches the x axis) of the graph y = (x - 4)^2


add it together and u should get ur answer. If I misinterpreted ur question, my bad

 

[Squar3root]

do you mean this area?

 

[Squar3root]

find the point of intersection of the 2 curves by solving simatenously

you should get x=1 (ignore x=12)

now you want the intergral of y=5x+4 from 0 to 1 PLUS the intergral of (x-4)^2 from 1 to 4 and that will give you your answer

 

[Smile12345]

So the area bounded by the two graphs, the x and y axis?

If so...

Find the point of intersection if u solve 5x + 4 = (x - 4)^2

U take the smaller x coordinate, since u can visually see that it would make sense if u graph it.

I got x = 1.

Now if u graphed it, draw a line down the point of intersection.

Find the Integral between 0 and 1 of the line y = 5x + 4

then find the integral between 1 and 4 (since that's where the parabola touches the x axis) of the graph y = (x - 4)^2


add it together and u should get ur answer. If I misinterpreted ur question, my bad

Yup, that's exactly what I meant...

 

[Smile12345]

find the point of intersection of the 2 curves by solving simatenously

you should get x=1 (ignore x=12)

now you want the intergral of y=5x+4 from 0 to 1 PLUS the intergral of (x-4)^2 from 1 to 4 and that will give you your answer

Yeah ok... I got 9!! Will have to check...

 

[Smile12345]

do you mean this area?

Yes. (roughly looks like that).

 

[Study notes]

ok cool, sorry I'm not in the mood to do any actual working out hope that's the answer

 

[Squar3root]

Yeah ok... I got 9!! Will have to check...

hmmm, i got 15.5

 

[Smile12345]

hmmm, i got 15.5

You're right... First part = 6.5??

 

[Smile12345]

You're right... First part = 6.5??

Got the next part as 9.... Sorry !

How about finding the area between the curves y = 2x^2 and y = x^2 -13x + 30

I've found the point of intersection is 2 4/13 is this right?!?!

Then do the integral from 0 to 2 4/13 with y = 2x^2 and add the integral from 2 4/13 to 3 (Think this is right.)

 

[Study notes]

the point of intersection for the x value (that you use for this) is x = 2 so yeah u're right.

not sure what u mean by 4/13 but yeah integral from 0 to 2 for y = 2x^2

Then integral from 2 to 3 for the graph y = x^2 - 13x + 30


then add them two like before. Same method as the last question, just the extra step of finding where it the second graph cuts the x axis at.