Evaluate definite integral. x if x<1; 1/x if x> or equal to 1.

[Lo.Lee.Ta.]

Evaluate definite integral. "x if x<1; 1/x if x> or equal to 1."

1. Consider the function: f(x) = {x if x<1
{1/x if x≥1

Evaluate the definite integral.

∫from 0 to 4 of f(x)dx


2. Okay, I think I vaguely remember something about these sorts of problems... Isn't it something like you choose which value to go along with depending on what the limits of the integral are?
But the limit 0 goes along with "x if x<1" and 1 through 4 go along with "1/x if x≥1."
And integrals just evaluate area, so can we break this up into 2 integral problems?

Integral #1

∫0 to 0
Wait...But that would be just zero. If the limits were from 0 to 0, then there'd be no area! :/

Integral #2

∫1 to 4 of (1/x)dx

= ln(4) - ln(1) = 1.386

Yeah, this is not the right answer.
I don't even really know what I'm doing... =_=
I'm probably not even going about it right at all...
Help?
Thank you SO much!

 

[haruspex]

Right approach, but where did you get "0 to 0" from in the first integral?

 

[Lo.Lee.Ta.]

I was thinking that the "x if x<1" meant that 1 could not be part of the limit...

But maybe it's okay to have the integral from zero to one because it's just the area right up to one?

So would it be instead:

Integral #1

∫0 to 1 of x

= (x^2)/2 |0 to 1

= (1^2)/2 - (0^2)/2

= 1/2 - 0 = 1/2

Integral #2

∫1 to 4 of (1/x)dx

= ln(4) - ln(1) = 1.386

So would the answer be .5 + 1.386 = 1.886 ?
Not sure at all... :/

 

[Ray Vickson]

I was thinking that the "x if x<1" meant that 1 could not be part of the limit...

But maybe it's okay to have the integral from zero to one because it's just the area right up to one?

So would it be instead:

Integral #1

∫0 to 1 of x

= (x^2)/2 |0 to 1

= (1^2)/2 - (0^2)/2

= 1/2 - 0 = 1/2

Integral #2

∫1 to 4 of (1/x)dx

= ln(4) - ln(1) = 1.386

So would the answer be .5 + 1.386 = 1.886 ?
Not sure at all... :/

The inequality x < 1 just means that '1' is excluded, but everything to the left of 1 is included. The area under the curve AT x = 1 is zero (that is, it is the area of a line segment, which = 0). So, yes: the first part is the integral from 0 to 1.

 

[SammyS]

I was thinking that the "x if x<1" meant that 1 could not be part of the limit...

But maybe it's okay to have the integral from zero to one because it's just the area right up to one?
...

If you want to get picky, for Integral#1 you could do

\displaystyle \lim_{\,a\to1^-\,}\int_0^a x\,dx\ .

But that will give the same result as \displaystyle \ \ \int_0^1 x\,dx\ .

 

[HallsofIvy]

I was thinking that the "x if x<1" meant that 1 could not be part of the limit...

Strictly speaking, as SammyS said, you can't. You need to use the limit as an improper integral. But even so, surely you realize that there are numbers between 0 and 1?

But maybe it's okay to have the integral from zero to one because it's just the area right up to one?

So would it be instead:

Integral #1

∫0 to 1 of x

= (x^2)/2 |0 to 1

= (1^2)/2 - (0^2)/2

= 1/2 - 0 = 1/2

Integral #2

∫1 to 4 of (1/x)dx

= ln(4) - ln(1) = 1.386

So would the answer be .5 + 1.386 = 1.886 ?
Not sure at all... :/

 

[Lo.Lee.Ta.]

Okay, so it's okay to have the limits from 0 to 1 and from 1 to 4.

But my homework computer program does not count the answer 1.886 as correct...

Did I make some mistake in calculating the integrals?
Do you have any idea where I went wrong?

Thank you!

 

[SammyS]

Okay, so it's okay to have the limits from 0 to 1 and from 1 to 4.

But my homework computer program does not count the answer 1.886 as correct...

Did I make some mistake in calculating the integrals?
Do you have any idea where I went wrong?

Thank you!

Maybe it wants the exact number, i.e. (1/2)+ln(4) .

 

[Lo.Lee.Ta.]

@SammyS - Wow! You were exactly right! Thank you so much! :D