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## Homework Statement

I am getting the answer as -(pi)

The answer given is (pi). There might be a sign error. I checked it many times but cannot find my fault. I need some help.

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- Thread starter zorro
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- #1

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I am getting the answer as -(pi)

The answer given is (pi). There might be a sign error. I checked it many times but cannot find my fault. I need some help.

- #2

Mark44

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The quantities in the two radicals are equal, so I don't see a problem there. One problem that I see that you might not have addressed is that the integral you started with is improper - the integrand is undefined at both x = a and x = b.## Homework Statement

I am getting the answer as -(pi)

The answer given is (pi). There might be a sign error. I checked it many times but cannot find my fault. I need some help.

## Homework Statement

## Homework Equations

## The Attempt at a Solution

- #3

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One problem that I see that you might not have addressed is that the integral you started with is improper - the integrand is undefined at both x = a and x = b.

Then that should lead to an indeterminable integral. Why there is only a difference in sign?

How do I proceed the correct way?

- #4

Mark44

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Not necessarily. You need to split the original integral into two integrals, and use limits to evaluate each.Then that should lead to an indeterminable integral.

[tex]\lim_{p \to a^+}\int_p^1 \frac{dx}{\sqrt{(x - a)(b - x)}} + \lim_{q \to b^-}\int_1^q \frac{dx}{\sqrt{(x - a)(b - x)}} [/tex]

I have no way of knowing, since I don't know what you did.Why there is only a difference in sign?

See above.How do I proceed the correct way?

- #5

Char. Limit

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- #6

Mark44

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- #7

jgens

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- #8

Char. Limit

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Yes, you're right of course. Don't know what I was thinking there.

- #9

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[tex]\lim_{p \to a^+}\int_p^1 \frac{dx}{\sqrt{(x - a)(b - x)}} + \lim_{q \to b^-}\int_1^q \frac{dx}{\sqrt{(x - a)(b - x)}} [/tex]

I am getting the first integral as (after applying limit)

Is it right?

- #10

Mark44

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When I evaluated the first integral I got -[sqrt(1-a)sqrt(b-1)tan^(-1)((a+b-2)/2sqrt(1-a)sqrt(b-1))] / [sqrt(1-a)(b-a)] + pi/2.

I suspect that the second integral ends up wiping out the first term above + another pi/2.

- #11

hunt_mat

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Note

[tex]

\int_{a}^{b}\frac{dx}{\sqrt{(x-a)(b-x)}}=\int_{a}^{b}\frac{dx}{\sqrt{\left(\frac{b-a}{2}\right)^{2} -\left( x-\frac{a+b}{2}\right)^{2}}}

[/tex]

Use the substitution:

[tex]

y=x-\frac{a+b}{2}

[/tex]

To obtain the integral:

[tex]

\int_{-\alpha}^{\alpha}\frac{dy}{\sqrt{\alpha^{2}-y^{2}}}

[/tex]

Then use the substitution [tex]y=\alpha\sin z[/tex] to get the answer quoted. I will leave you to fill in the details and find out what alpha is.

- #12

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Note

[tex]

\int_{a}^{b}\frac{dx}{\sqrt{(x-a)(b-x)}}=\int_{a}^{b}\frac{dx}{\sqrt{\left(\frac{b-a}{2}\right)^{2} -\left( x-\frac{a+b}{2}\right)^{2}}}

[/tex]

Use the substitution:

[tex]

y=x-\frac{a+b}{2}

[/tex]

To obtain the integral:

[tex]

\int_{-\alpha}^{\alpha}\frac{dy}{\sqrt{\alpha^{2}-y^{2}}}

[/tex]

Then use the substitution [tex]y=\alpha\sin z[/tex] to get the answer quoted. I will leave you to fill in the details and find out what alpha is.

This is much better.

But I see no use of restricting the domain other than our own satisfaction. I mean we don't use that condition anywhere in the solution later.

I found out that there is a change in sign only because of taking {(a-b)^2}/2 as I did instead of {(b-a)^2}/2 as you did.

The formula for

Is there any change to be made in this formula such as applying mod. function so that I can use it in both the cases {(a-b)^2}/2 & {(b-a)^2}/2 which are equivalent algebraically but not in calculus?

- #13

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^{-1}part? I used wolfram alfpha, and the indefinite integral involves the inverse tangent function.

When I evaluated the first integral I got -[sqrt(1-a)sqrt(b-1)tan^(-1)((a+b-2)/2sqrt(1-a)sqrt(b-1))] / [sqrt(1-a)(b-a)] + pi/2.

I suspect that the second integral ends up wiping out the first term above + another pi/2.

Did you miss the root? How can you get an arctan function with a root?

I applied the formula I gave in the previous post and got arcsin.

- #14

hunt_mat

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Technically what we're doing is taking away what is known as a null set {a,b}, this will have no effect whatsoever on the value of the integral. The signs should all cancel out in the end.

Mat

- #15

Mark44

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No, I didn't omit the square root. This is what I entered into wolframalpha:Did you miss the root? How can you get an arctan function with a root?

integrate 1/sqrt((x-a)(b-x)) dx

The result was what I posted earlier in this thread.

It is often the case that integrating by two different techniques gives different antiderivatives, but they can differ at most by a constant.I applied the formula I gave in the previous post and got arcsin.

Last edited:

- #16

Mark44

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At some point you are using the Fundamental Theorem of Calculus to find an antiderivative, which is then evaluated at the two endpoints of the interval. The FTC requires the integrand to be continuous on the closed interval [a, b], which means it must be also defined at all points in the interval, including the two endpoints. This is the reason for the need to work with limits at both endpoints in this problem. The fact that we're dealing with a set of measure zero is irrelevant.As the integral was from a to b, I made the assumption that b>a, and so the term (b-a).2>0.

Technically what we're doing is taking away what is known as a null set {a,b}, this will have no effect whatsoever on the value of the integral. The signs should all cancel out in the end.

- #17

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No, I didn't omit the square root. This is what I entered into wolframalpha:

integrate 1/sqrt((x-a)(b-x)) dx

The result was what I posted earlier in this thread.

It is often the case that integrating by two different techniques gives different antiderivatives, but they can differ at most by a constant.

I entered the function there and got it as -

[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP258319ddgi17bga1e8hh000021dhe85848hb983f?MSPStoreType=image/gif&s=33&w=483&h=62 [Broken]

How did you evaluate the limit? Is it not too complex? Why are the like terms in numerator and denominator not cancelled here in wolframalpha?

Last edited by a moderator:

- #18

vela

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I suspect the mistake you made was saying [itex]\sqrt{c^2}=c[/itex] when you should actually say [itex]\sqrt{c^2}=|c|[/itex]. It's not that calculus and algebra follow different rules. You just need to be more careful about simplifying expressions.This is much better.

But I see no use of restricting the domain other than our own satisfaction. I mean we don't use that condition anywhere in the solution later.

I found out that there is a change in sign only because of taking {(a-b)^2}/2 as I did instead of {(b-a)^2}/2 as you did.

The formula for

Is there any change to be made in this formula such as applying mod. function so that I can use it in both the cases {(a-b)^2}/2 & {(b-a)^2}/2 which are equivalent algebraically but not in calculus?

- #19

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At some point you are using the Fundamental Theorem of Calculus to find an antiderivative, which is then evaluated at the two endpoints of the interval. The FTC requires the integrand to be continuous on the closed interval [a, b], which means it must be also defined at all points in the interval, including the two endpoints. This is the reason for the need to work with limits at both endpoints in this problem. The fact that we're dealing with a set of measure zero is irrelevant.

I did not get one thing....see this example-

Here we can divide numerator and denominator by cos

- #20

Mark44

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In any case, this 1/(1 + cos^2(x)) is defined on and continuous on [0, pi/2], so this is different from the integral in the OP.

If you are evaluating an integral such as this:

[tex]\int_a^b f(x)dx[/tex]

the integrand

- #21

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It's not obvious to me that dividing numerator and denominator by cos^2(x) leads to an integral that can be evaluated easily, since the integrand becomes sec^2(x)/(sec^2(x) + 1).

You can write the denominator as 2 + tan

In any case, this 1/(1 + cos^2(x)) is defined on and continuous on [0, pi/2], so this is different from the integral in the OP.

Yes this is different but I found some relevancy.

- #22

Mark44

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The new integral

[tex]\int_0^{\pi/2}\frac{sec^2(x)dx}{sec^2(x) + 2}[/tex]

is an improper integral, since the integrand is undefined at x = pi/2. To evaluate it you would need to use a limit, like so:

[tex]\lim_{b \to \pi/2^-}\int_0^b\frac{sec^2(x)dx}{sec^2(x) + 2}[/tex]

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- #24

Mark44

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[tex]\int_0^{\pi/2}\frac{sec^2(x)dx}{sec^2(x) + 2}[/tex]

without using limits, it's not doing things the right way.

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Thanks Mark!

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