Evaluate.....

Question

Evaluate.....

anonymous · Answer

ok

anonymous · Answer

$$\int\limits_{5}^{0} 2x \sqrt{x+4} dx$$

anonymous · Answer

use substitution

zepdrix · Answer

Having a little trouble with the 2x term marnie? :)

anonymous · Answer

u=x+4
du=dx
x=u-4

=2(u-4)(u) du?

anonymous · Answer

You don't want any multiplication between variables!!

zepdrix · Answer

$$\huge =2(u-4)(\sqrt u)du$$I think you missed one small detail marnie :O

anonymous · Answer

oh yes i missed that....ok im honestly lost from here.....if i can't have multiplication between variables what do i do?

zepdrix · Answer

Well now your multiplication isn't so bad, since the squareroot doesn't have a bunch of funky junk inside of it.
Rewrite the sqrt(u) using fractional exponents, then simply expand out the brackets! :)

anonymous · Answer

ok:

$$\int\limits_{0}^{5}u ^{3/2}-4u ^{1/2)}$$

zepdrix · Answer

$$\large 2\int\limits (u-4)\sqrt u du \quad = \quad 2\int\limits (u-4)u^{1/2} du$$
$$\large  =2\int\limits u^{1+1/2}-4u^{1/2} du \quad \rightarrow \quad 2\int\limits u^{3/2}-4u^{1/2}du$$
Yes looks good ^^ I'm not sure where your 2 went though heh

anonymous · Answer

oh yes
ok....

2 * (2/3u^3/2-8/3u^3/2)

=-4u^3/2

anonymous · Answer

=-4(x+4)^3/2

anonymous · Answer

Oh no.....i think that's wrong......i forgot about my 0 and 5. ahh confusing

zepdrix · Answer

On your second term, it doesn't look like you increased the power by 1.
should be u^5/2 i believe. So they won't combine nicely.

anonymous · Answer

oh yes ur right

anonymous · Answer

=4/5u^5/2 - 16u^3/2


then what do i do? Do i enter my 0 and 5?

zepdrix · Answer

The second term has a coefficient of 16/3, the way you had it before :)

Recall that the limits of integration were for our X variable!!
x=0 and x=5 are our limits of integration.

We have 2 options from here,

We can either rewrite our solution in terms of x, undoing the U substition we made earlier, and then we can plug in the 0 and 5.

OR

We can find new limits of integration by looking at our substitution, and plugging in the 0 and 5.

They both will take about the same amount of work, so which ever way you prefer.

anonymous · Answer

ok phew! That makes sense. Thank u!

zepdrix · Answer

Yay team \:D/

anonymous · Answer

too bad i can't bring you to my exam hahah

zepdrix · Answer

lol XD

anonymous · Answer

actually I have a question....I have the answer to this question, but it's different and i noticed this other person changed the 0 and 5, to a 9 and 4. Why would they do that?