Find ∫ 1/(1-4x^2) dx with a low limit of 0 and a high limit of 1

**bookworm247** · August 23rd 2010, 02:54 PM

Find ∫ 1/(1-4x^2) dx with a low limit of 0 and a high limit of 1 (another way to put it is Integrate[1/(1 - 4 x^2), {x, 0, 1}]

Anyway this is what I did:

I knew that this definite integral was = 1/2 tan^-1 (2) which equal about 1.553574...

I got the above from knowing that the integral of 1/(4x^2 )+1 is 1/2 tan ^-1 (2x)
=1/2 tan ^-1 (2x) + constant

(what I want to know is if this is enough work/steps to show how I got the answer- if it is not could someone tell what steps I would have to show- Thanks for the help)

**Plato** · August 23rd 2010, 03:06 PM

Use partial fractions $\dfrac{1}{{1 - 4x^2 }} = \dfrac{1}{{2(2x + 1)}} - \dfrac{1}{{2\left( {2x - 1)} \right)}}$ .

**bookworm247** · August 23rd 2010, 03:27 PM

Sorry I wrote the wrong thing- it should be a + instead of a -

∫ 1/(1+4x^2) dx with a low limit of 0 and a high limit of 1

Thanks for the help and sorry for the mix-up

**Ackbeet** · August 23rd 2010, 05:37 PM

Right. That sign difference is critical. Otherwise, the function blows up in the middle of the interval, and doesn't converge!

The amount of work you've shown might or might not be enough. It depends on whether you're allowed to just "know" that the antiderivative is the inverse tangent, or whether you're supposed to derive it using a trig substitution.

Incidentally, your numerical value for the integral is off by precisely 1.

**bookworm247** · August 24th 2010, 02:01 PM

So how would you go about solving this problem (becasue I did know that the antiderivative is the inverse tangent but I don't iknow if that is enough information- and I was off my one beacsue that was a typo sorry)

I know that since 1/2 tan^-1 (2) which equal about 0.553574...
and that 1/2 tan^-1 (2) can also be written as (1/2) ArcTan[2] so then would I use
∫ arctan x dx = x arctan x - (1/2) ln(1+x2) + C

So I would do

1/2 ∫ arctan 2 dx = 1/2 (2 arctan (2)-(1/2) ln(1+2(2)) + C

(am I doing this right or am I completely wrong- if so could some show me the steps I neeed to take to finish this or what I am doing wrong)

Thank you for the help I appreciate it.

**Ted** · August 24th 2010, 05:07 PM

Re-write the integral to get:

$\frac{1}{4} \, \int \dfrac{dx}{\frac{1}{4}+x^2}$

Now, use the following formula:

$\int \dfrac{dx}{x^2+a^2}=\dfrac{1}{a} arctan\left(\dfrac{x}{a}\right) + C$ for $a \neq 0$ .

**bookworm247** · August 25th 2010, 02:03 PM

So would this be the answer:
the a would be the 1/4
1/(1/4) arctan (x/(1/4) +C = 4 arctan (4x) +C

But now what do I do, I know that it has a low limit of 0 and a high limit of 1 so what do I do with these numbers to complete the problem/get the answer?

Thanks for the help

**bugatti79** · August 25th 2010, 02:52 PM

Originally Posted by bookworm247

So would this be the answer:
the a would be the 1/4
1/(1/4) arctan (x/(1/4) +C = 4 arctan (4x) +C

But now what do I do, I know that it has a low limit of 0 and a high limit of 1 so what do I do with these numbers to complete the problem/get the answer?

Thanks for the help

No, $a^2=\frac{1}{4}$ therefore $a=\frac{1}{2}$ because

$\sqrt{a^2}=\sqrt{\frac{1}{4}}$ therefore $a=\frac{\sqrt{1}}{\sqrt{4}}=\frac{1}{2}$

**bookworm247** · August 25th 2010, 06:01 PM

So it would be
1/(1/2) arctan (x/(1/2) +C
= 2 arctan (2x) +C

However I am still confused on what to do with the lower and higher limits of 0 and 1 (could someone show me?)

Thanks

**bugatti79** · August 26th 2010, 01:53 AM

Originally Posted by bookworm247

So it would be
1/(1/2) arctan (x/(1/2) +C
= 2 arctan (2x) +C

However I am still confused on what to do with the lower and higher limits of 0 and 1 (could someone show me?)

Thanks

Your answer will be = 1/4[(2 arctan (2(1)) +C)- (2 arctan (2(0)) +C)]
becasue it is a definite integral ie, you were given upper and lower limits then your c's will cancel out.

**bookworm247** · August 26th 2010, 09:30 AM

Thank you for the help.

Since the C will cancel out then the final answer should be this correct:

1/4[(2 arctan (2(1)) +C)- (2 arctan (2(0)) +C)]
=1/4[(2 arctan (2) - 2 arctan (0)]
=1/4 [2 arctan (2) - 0
= 1/4 [2 arctan (2)]
= 1/2 arctan (2) This is the final solution correct?

Thank you for all the help

**bugatti79** · August 26th 2010, 10:25 AM

Yes, I believe so.

**HallsofIvy** · August 26th 2010, 10:59 AM

Originally Posted by bookworm247

Find ∫ 1/(1-4x^2) dx with a low limit of 0 and a high limit of 1 (another way to put it is Integrate[1/(1 - 4 x^2), {x, 0, 1}]

Anyway this is what I did:

I knew that this definite integral was = 1/2 tan^-1 (2) which equal about 1.553574...

I got the above from knowing that the integral of 1/(4x^2 )+1 is 1/2 tan ^-1 (2x)
=1/2 tan ^-1 (2x) + constant

(what I want to know is if this is enough work/steps to show how I got the answer- if it is not could someone tell what steps I would have to show- Thanks for the help)

Okay, so the problem is actually $\int_0^1 \frac{dx}{1+ 4x^2}$ .
Yes, it is true that $\int \frac{dx}{1+ x^2}= \frac{1}{2}tan^{-1}(2x)+ C$ . Now, evaluate that between 0 and 1:
$\frac{1}{2}(tan^{-1}(2)- tan(0))$
[tex]tan^{-1}(0)= 0[/math, of course, but you will need a calculator to find $tan^{-1}(2)$ . Be sure to put your calculator in "radian" mode.

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