Integration Question (1 Viewer)

Sanjeet

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This was a multiple choice question in my trials I just did, anyone know how to do this? I thought it would be 8, but I guessed 18
 
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deswa1

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Well, you can split the integral to f(x-3) and 2. We know that the integral from 3 to 7 of f(x-3) equals the integral of f(x) from 0 to which equals 6. The integral of 2 from 3 to 7 is 2(7)-2(3)=8. Sum the two and you get 14.

This might be hard to visualise- I can latex it later if no one does it soon
 

Sanjeet

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Well, you can split the integral to f(x-3) and 2. We know that the integral from 3 to 7 of f(x-3) equals the integral of f(x) from 0 to which equals 6. The integral of 2 from 3 to 7 is 2(7)-2(3)=8. Sum the two and you get 14.

This might be hard to visualise- I can latex it later if no one does it soon
yeah but the 2 is included in the function - it's not outside it
 

Carrotsticks

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yeah but the 2 is included in the function - it's not outside it
Oh I didn't see that, that's a bit silly then because (x-3)+2 is just (x-1). Would be much more interesting if 2 was outside because you have to realise you're just adding a 2x4 rectangle.

Also, I'm not entirely happy with the question and I have a feeling it is invalid due to assumptions. Will post up a pic in a few minutes.
 

Sanjeet

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Oh I didn't see that, that's a bit silly then because (x-3)+2 is just (x-1). Would be much more interesting if 2 was outside because you have to realise you're just adding a 2x4 rectangle.

Also, I'm not entirely happy with the question and I have a feeling it is invalid due to assumptions. Will post up a pic in a few minutes.
Yea that's exactly what I thought, so I thought it'd just become and I have no idea how to do that
 

seanieg89

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I am pretty sure those square brackets should not exist. Without them the question makes perfect sense and is of an appropriate level of difficulty for MC.

As the OP has written the question, the question does not make sense. There is no reason why the domain of f has to include anything outside of the interval [0,4].

Eg by multiplying the function sqrt(x(4-x)) by a suitable constant we can obtain a function which satisfies the hypotheses of the question but such that the latter integral is not even defined.
 

Sanjeet

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I'm pretty sure they made a mistake, I copied down exactly what was written in the exam. I'll let you guys know when we get the results
 

Sanjeet

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Here's the question just so you know I didn't make an error:
IMG_0640[1].jpg
My teacher is going to discuss it with the other teachers so I will find out by monday
 

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