# Thread: Difficult definite integral

1. ## Difficult definite integral

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.

2. 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.

3. 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

4. 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

$\chi$ $\sigma$

5. 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?

6. 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

7. 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.