Does this work and if so how?

Ok, so i'm wondering if this will work and if so how? Can 4 be used for both parts because when we put it into sin we are assuming it is 4 radians and when we put it into the second part we are assuming it is 4 as an integer. Does this work?

Original post by CIEBioloysifh

Ok, so i'm wondering if this will work and if so how? Can 4 be used for both parts because when we put it into sin we are assuming it is 4 radians and when we put it into the second part we are assuming it is 4 as an integer. Does this work?

That's correct. You can use www.wolframalpha.com to check work like this by the way.

http://www.wolframalpha.com/input/?i=integrate+(cos+x+-+3%2Fx%5E5)+dx+between+x+%3D+1+and+x+%3D+4

(edited 7 years ago)

Reply 2

CIEBioloysifh

OP

13

Original post by Mr M

That;s correct. You can use www.wolframalpha.com to check work like this by the way.

I know it's correct, however, I came to TSR looking for a clear answer as to how this is the case as I'm really struggling to explain it to my friend. It's not the integration, it's the fact that plugging the value into the integral we use 4 as 4 radians and 4 as an integer in the second part.

Original post by CIEBioloysifh

I know it's correct, however, I came to TSR looking for a clear answer as to how this is the case as I'm really struggling to explain it to my friend. It's not the integration, it's the fact that plugging the value into the integral we use 4 as 4 radians and 4 as an integer in the second part.

If that's bugging you for some reason (and I don't really understand why it should) then treat this as two separate integrals. You'll get the same answer.

\displaystyle \int_1^4 \cos x \, dx - \int_1^4 \frac{3}{x^5} \, dx

Reply 4

CIEBioloysifh

OP

13

Original post by Mr M

If that's bugging you for some reason (and I don't really understand why it should) then treat this as two separate integrals. You'll get the same answer.

\displaystyle \int_1^4 \cos x \, dx - \int_1^4 \frac{3}{x^5} \, dx

He's telling me that in the first section we are plugging in 4 radians and in the second section we're plugging in the integer 4, therefore the numbers we are plugging in aren't the same and they aren't the same limit for both terms because we are plugging in x= 4 radians

Reply 5

atsruser

11

Original post by CIEBioloysifh

He's telling me that in the first section we are plugging in 4 radians and in the second section we're plugging in the integer 4, therefore the numbers we are plugging in aren't the same and they aren't the same limit for both terms because we are plugging in x= 4 radians

The radian measure of an angle is given by the ratio of two lengths

s/r

so is unitless, since the metre units in the top and bottom cancel. A radian is therefore a "pure" number, and 4 radians is thus simply the integer 4, it's not 4 metres or 4 seconds or 4 "radian units", or whatever. You only know that 4 means 4 radians from the context in which it is used.

Does that explain it?

Original post by CIEBioloysifh

He's telling me that in the first section we are plugging in 4 radians and in the second section we're plugging in the integer 4, therefore the numbers we are plugging in aren't the same and they aren't the same limit for both terms because we are plugging in x= 4 radians

Radians are dimensionless so there really isn't a problem here. Just tell your friend to accept it. If he doesn't, he is certainly going to struggle to accept integration by substitution involving a trigonometric substitution!

Edit: atruser's explanation is better than mine.

(edited 7 years ago)

Reply 7

CIEBioloysifh

OP

13

Original post by atsruser

The radian measure of an angle is given by the ratio of two lengths

s/r

so is unitless, since the metre units in the top and bottom cancel. A radian is therefore a "pure" number, and 4 radians is thus simply the integer 4, it's not 4 metres or 4 seconds or 4 "radian units", or whatever. You only know that 4 means 4 radians from the context in which it is used.

Does that explain it?

Yes thank you, that made it very clear to me, however, no luck with my friend still refusing to accept that the limit of the integral can be used this way.

Original post by CIEBioloysifh

Yes thank you, that made it very clear to me, however, no luck with my friend still refusing to accept that the limit of the integral can be used this way.

Does your friend also believe the Earth is flat? Mathematicians have been integrating happily for 350 years!

Reply 9

atsruser

11

Original post by CIEBioloysifh

Yes thank you, that made it very clear to me, however, no luck with my friend still refusing to accept that the limit of the integral can be used this way.

It's not clear to me what he doesn't like, but here's another argument:

Consider two sin functions, one of which takes a radian argument (i.e. the usual calculus type of sin function) and one of which takes a degree argument. These are different functions because their domains differ e.g. we could have:

\sin^{c} : [0,2\pi] \to [-1,1]

\sin^{\circ} : [0,360] \to [-1,1]

We can make the second out of the first by function composition. Let

f(x) = \frac{\pi x}{180}

then

\sin^{\circ} = \sin^{c} \circ f

and

\sin^{\circ} x = \sin^{c}(\frac{\pi x}{180})

Now we can write the integral

\int_0^4 \sin^c x +\sin^{\circ} x \ dx

It should now be clear(?) that:

a) this integral makes sense, since we can write it all in terms of

\sin^{c}

if we want.
b) the limit 4 is neither a radian nor a degree, since when we feed it to the first sin function, it works as a radian, and when we feed it to the second, it works as a degree.

[edit: of course, I got this wrong, since we actually feed the 4 into the integrated sin functions, namely the corresponding cos functions, but the point still stands]

(edited 7 years ago)

Does this work and if so how?

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