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Definite integration- Evaluate

Question:
Given that ∫ upper limit 9 lower limit 3 f(x)dx=7, evaluate∫ upper limit 9 lower limit 3 2f(x)+1dx.

As far as I understand
∫ upper limit 9 lower limit 3 f(x)dx = F(9)-F(3)
So, F(9)-F(3)=7

If I knew the f(x) I could solve it easy enough.
If d/dx F(x)=f(x)

Help in the right direction would be grateful.

Thank you
Reply 1
Avatar for mqb2766
mqb2766
21
Original post by KingRich
Question:
Given that ∫ upper limit 9 lower limit 3 f(x)dx=7, evaluate∫ upper limit 9 lower limit 3 2f(x)+1dx.

As far as I understand
∫ upper limit 9 lower limit 3 f(x)dx = F(9)-F(3)
So, F(9)-F(3)=7

If I knew the f(x) I could solve it easy enough.
If d/dx F(x)=f(x)

Help in the right direction would be grateful.

Thank you

A rough sketch of f(x) might help. Keep it simple, you can choose what the "curve" looks like. The area under it (between the curve and the x-axis) is 7.
Then double f(x) and add 1 to 2f(x), doing another sketch of 2f(x)+1. How will (the area under) the curve change?

While its simple to do algebraically, having a proper insight is useful.
Reply 2
Avatar for KingRich
KingRich
OP
14
Original post by mqb2766
A rough sketch of f(x) might help. Keep it simple, you can choose what the "curve" looks like. The area under it (between the curve and the x-axis) is 7.
Then double f(x) and add 1 to 2f(x), doing another sketch of 2f(x)+1. How will (the area under) the curve change?

While its simple to do algebraically, having a proper insight is useful.

I did try to visualise as you said but if what I believe you mean by double f(x) is, then I end up with f(6)-f(18).

Erm, am I right that a transformation in the graph here is happening? For example the graph is vertically stretching by 2?
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by KingRich
I did try to visualise as you said but if what I believe you mean by double f(x) is, then I end up with f(6)-f(18).

Erm, am I right that a transformation in the graph here is happening? For example the graph is vertically stretching by 2?

First paragraph - thats incorrect, its not a horizontal (x-axis) scaling.
Second paragraph - thats right, its a vertical scaling (y-axis) by 2 and add 1.

Tbh, I'd change the question slightly. If the definite integral is 6 (rather than 7) between x=3 and x=9, what is the simplest function f(x) that you could imagine which has that area between those limits. Then multiply it (y-axis) by 2 and add 1. What is the new area?

Then it should be easy to work out if the original definite integral is 7.
(edited 2 years ago)
Reply 4
Avatar for KingRich
KingRich
OP
14
Original post by mqb2766
First paragraph - thats incorrect, its not a horizontal (x-axis) scaling.
Second paragraph - thats right, its a vertical scaling (y-axis) by 2 and add 1.

Tbh, I'd change the question slightly. If the definite integral is 6 (rather than 7) between x=3 and x=9, what is the simplest function f(x) that you could imagine which has that area between those limits. Then multiply it (y-axis) by 2 and add 1. What is the new area?

Then it should be easy to work out if the original definite integral is 7.

Sorry, I am so lost right now. I might have to take a break and come back
Reply 5
Avatar for mqb2766
mqb2766
21
Original post by KingRich
Sorry, I am so lost right now. I might have to take a break and come back

You may be overthinking it a bit. Consider
f(x) = 1
What is the area of the (rectangle) between x=3 and x=9 (definite integral)?

Then what is the area under function g(x) (between it and the x-axis) where
g(x) = 2f(x) + 1 = f(x) + f(x) + 1 = ...
again between x=3 and x=9.

Try sketching what you think and upload?
(edited 2 years ago)
Reply 6
Avatar for KingRich
KingRich
OP
14
Original post by mqb2766
You may be overthinking it a bit. Consider
f(x) = 1
What is the area of the (rectangle) between x=3 and x=9 (definite integral)?

Then what is the area under function g(x) (between it and the x-axis) where
g(x) = 2f(x) + 1 = f(x) + f(x) + 1 = ...
again between x=3 and x=9.

Try sketching what you think and upload?

I believe our initial approach on this was wrong.

The 1 is yet to be integrated. The example given tells us that between the limits 9 and 3 our y=7 and upon integration of 1 between those limits you find the value of 6.

So, 2f(x)=14 + 6= 20

4AE6F93C-05F5-4B3B-8378-C7C7AF777C1F.jpeg

I think the way that it has been written is misleading, or maybe a lack of understanding has confused me.
I realised now a transformation cannot happen between two points only. If there’s a transformation of any value it effects the whole graph.
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by KingRich
I believe our initial approach on this was wrong.

The 1 is yet to be integrated. The example given tells us that between the limits 9 and 3 our y=7 and upon integration of 1 between those limits you find the value of 6.

So, 2f(x)=14 + 6= 20

4AE6F93C-05F5-4B3B-8378-C7C7AF777C1F.jpeg

I think the way that it has been written is misleading, or maybe a lack of understanding has confused me.
I realised now a transformation cannot happen between two points only. If there’s a transformation of any value it effects the whole graph.

That looks good. You're simply scaling and translating the y axis so the area under it has a similar transformation
2f(x)
is
2*area under f()
and adding 1 such as
2f(x) + 1
means you add a strip of height 1 and width 6 (9-3) to the function, so the area is increased by 6.

As you say
[br]∫392f(x)+1dx=2∫39f(x)dx+∫391dx=2∗7+6=20[br][br]\int_3^9 2f(x)+1 dx = 2\int_3^9 f(x) dx + \int_3^9 1 dx = 2*7 + 6 = 20[br]
but think about the transformation to the function and hence the area under it.
(edited 2 years ago)

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