Find Values of a for Integral Calculus Homework

[planauts]

Homework Statement

Find the values of a > 0, such that $$\int_{a }^{a^2} \frac{1}{1+x^2} dx = 0.22$$

The Attempt at a Solution

I integrated it and got:

$$\int_{a }^{a^2} \frac{1}{1+x^2} dx = 0.22$$

$$arctan(a^2) - arctan(a) = 0.22$$

I am stuck after this, how would I solve this :-/

Thanks

 

[dextercioby]

Just apply <tangent> on both sides of your equation, use some trigonometry and see what you get

 

[planauts]

$$a^2 - a - tan(0.22) = 0$$
I got 1.188 and -0.188
The correct answer is a = 2.04, 2.62

 

[fzero]

You might want to look up (or prove) the correct identity for $$\tan (A-B)$$.

 

[planauts]

You might want to look up (or prove) the correct identity for $$\tan (A-B)$$.

I don't think that the tangent is the problem. I think my equation is wrong for some reason. Because when I graphed it the points don't make sense.

Do you see anything wrong with my math (integrating 1/(1+x^2) etc.)?

 

[dextercioby]

The integration is ok. The trigonometry is the issue.

 

[planauts]

The integration is ok. The trigonometry is the issue.

Then why doesn't the answer from the key work in the equation :-/

 

[fzero]

Then why doesn't the answer from the key work in the equation :-/

Are you using degrees instead of radians?

 

[dextercioby]

Let's try it with 2.04

Equation proofing is: arctan (2.04^2) = 76.488 deg. arctan 2.04 = 63.886 deg. Difference is 13.602 degrees = 0.23 radians.

So it's correct. The other solution can be checked for accuracy in the same manner.

 

[planauts]

Let's try it with 2.04

Equation proofing is: arctan (2.04^2) = 76.488 deg. arctan 2.04 = 63.886 deg. Difference is 13.602 degrees = 0.23 radians.

So it's correct. The other solution can be checked for accuracy in the same manner.

You are saying that $$arctan(a^2) - arctan(a) = 0.22$$ is the correct equation.

So that's good. But now the problem is manipulating the equation to get the value of a.
How would I do that?

You said:

Just apply <tangent> on both sides of your equation, use some trigonometry and see what you get

I think that is where I made the mistake...I have to apply <tangent> to the whole "left hand" side as a one. I split the tangent and distributed it. I don't think that is allowed.

So if I change it: $$tan[arctan(a^2) - arctan(a)] = tan(0.22)$$

Is this what fzero meant?

$$\frac{a^2 - a}{1 - a^2 * a} = \frac{a^2 - a}{1 - a^3} = 0.22$$

$$a^2 - a = 0.22 (1 - a^3)$$
$$0 = 0.22 - 0.22 a^3 - a^2 + a$$

I got a = 1 OR a = -5.3588 OR a = -0.2866

 

[fzero]

Is this what fzero meant?

Yes but

$$\tan(\theta_1 - \theta_2 ) = \frac{\tan \theta_1 - \tan\theta_2}{1 + \tan \theta_1 \tan\theta_2},$$

so you should find

$$\frac{a^2 - a}{1 \mathbf{+} a^2 * a} = \frac{a^2 - a}{1 \mathbf{+} a^3} = 0.22$$

 

[dextercioby]

Looks likes a nasty cubic one gets. Cardano's formulae should be used to solve it or Maple, Mathematica if you have them.

 

[planauts]

Yes but

$$\tan(\theta_1 - \theta_2 ) = \frac{\tan \theta_1 - \tan\theta_2}{1 + \tan \theta_1 \tan\theta_2},$$

so you should find

$$\frac{a^2 - a}{1 \mathbf{+} a^2 * a} = \frac{a^2 - a}{1 \mathbf{+} a^3} = 0.22$$

I got -0.184556093 and 1.946862606 = a, which still isn't right :(

 

[VietDao29]

Yes but

$$\tan(\theta_1 - \theta_2 ) = \frac{\tan \theta_1 - \tan\theta_2}{1 + \tan \theta_1 \tan\theta_2},$$

so you should find

$$\frac{a^2 - a}{1 \mathbf{+} a^2 * a} = \frac{a^2 - a}{1 \mathbf{+} a^3} = 0.22$$

I got -0.184556093 and 1.946862606 = a, which still isn't right :(

There's still a little mistake there. You are taking tangent value on both sides, remember? :)

So, it should read something like this:
$$\tan \left( \arctan(a ^ 2) - \arctan(a) \right) = \tan (0.22)$$
$$\Leftrightarrow \frac{a ^ 2 - a}{1 + a ^ 3} = \tan (0.22)$$.

Can you take it from here? :)

And, by the way, the book seems to have missed one possible value for a.

 

[planauts]

There's still a little mistake there. You are taking tangent value on both sides, remember? :)

So, it should read something like this:
$$\tan \left( \arctan(a ^ 2) - \arctan(a) \right) = \tan (0.22)$$
$$\Leftrightarrow \frac{a ^ 2 - a}{1 + a ^ 3} = \tan (0.22)$$.

Can you take it from here? :)

And, by the way, the book seems to have missed one possible value for a.

Oh my god! Thank you so much. I don't know why I wasn't seeing this...
I got 2.61737, 2.0416455 AND -0.223619421 as the roots.

Thanks to everyone who helped me, especially fzero :)

 

[VietDao29]

Oh my god! Thank you so much. I don't know why I wasn't seeing this...
I got 2.61737, 2.0416455 AND -0.223619421 as the roots.

Thanks to everyone who helped me, especially fzero :)

Yup, looks good to me.

However, since the problem just asks for positive values of a (a > 0), so you don't need to take a = -0.223619421 into consideration.