Difficult definite integral

**Glitch** · October 26th 2010, 04:17 AM

The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$

I'm stumped. Not sure how to attempt this.

Any assistance would be great.

**HappyJoe** · October 26th 2010, 04:23 AM

How about first taking $t^2$ from within the square root, and move it outside as $|t|$ :

$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$

where $|t|=-t$ , because we find ourselves in the range $-1\leq t\leq 0$ . This last expression can be more easily integrated. I suggest substitution.

**tonio** · October 26th 2010, 04:25 AM

Originally Posted by Glitch

The question:
$\int_{-1}^{0} \sqrt{t^2 + t^4} dt$

I'm stumped. Not sure how to attempt this.

Any assistance would be great.

The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.

Note: the last one is an automatic integral!

Tonio

**chisigma** · October 26th 2010, 05:26 AM

Originally Posted by tonio

The integrand function is even, so we can as well work with $\int\limits_0^1\sqrt{t^2+t^4}dt=\frac{1}{2}\int\li mits^1_0 2t\sqrt{1+t^2}dt =$ ....take it from here.

Note: the last one is an automatic integral!

Tonio

I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...

$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$

It is quite obvious that the second term of the 'identity' (1) is an odd function

...

Kind regards

$\chi$ $\sigma$

**Glitch** · October 26th 2010, 05:35 AM

Originally Posted by HappyJoe

How about first taking $t^2$ from within the square root, and move it outside as $|t|$ :

$\sqrt{t^2+t^4} = \sqrt{t^2(1+t^2)} = |t|\sqrt{1+t^2} = -t\sqrt{1+t^2},$

where $|t|=-t$ , because we find ourselves in the range $-1\leq t\leq 0$ . This last expression can be more easily integrated. I suggest substitution.

I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?

**tonio** · October 26th 2010, 06:53 AM

Originally Posted by chisigma

I wonder how You can say: the integrand function is even and immediately after [implicity] write the 'identity'...

$\displaystyle \sqrt{t^{2} + t^{4}} = t\ \sqrt{1+t^{2}}$

It is quite obvious that the second term of the 'identity' (1) is an odd function

...

Kind regards

$\chi$ $\sigma$

This really took me by surprise, though I think it doesn't matter here since the even thing was just to change the integral's limits.

In fact I should have writtent $\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}$ , but in this case is irrelevant since the

variable t is already positive in the new integration's limits.

Tonio

**TheCoffeeMachine** · October 26th 2010, 07:37 AM

Originally Posted by Glitch

I'm a little confused as to why we make |t| = -t. Is it because the area under the curve must be positive? i.e. y >= 0 for all t?

Let $-u = t$ , then $dt = -du$ , then:

$\displaystyle \int_{-1}^{0} \sqrt{t^2+t^4}\;{dt} = -\int_{1}^{0}\sqrt{(-u)^2+(-u)^4}\;{du} = -\int_{1}^{0}\sqrt{u^2+u^4}\;{du} = -\int_{1}^{0} \sqrt{t^2+t^4}\;{dt}.$

Now we can write $\sqrt{t^2+t^4} = t\sqrt{1+t^2}$ , as our limits are non-negative.

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