Need to find a shaded area between y = x^2-2x and the x-axis. Thanks!

Question

parthkohli · Answer

Hello. The integral of a function is the area between that and the x-axis.

anonymous · Answer

I need to set the two equations to get the integers. One I know the x-values I can sub them into my def integral. BUT...what is the y=? if the shaded area ends at the x-axis?

parthkohli · Answer

Err... I am not able to understand you.$$y  = x^2 - 2x $$essentially is the same as$$f(x)  = x^2 - 2x$$Also, this is not a definite integral because there are no upper or lower limits on the $x$. So you'll just have to find an expression for the area.

parthkohli · Answer

But I am sorry if I misunderstood your question...

parthkohli · Answer

Can you integrate $x^2 - 2x$?

anonymous · Answer

2x-2?

anonymous · Answer

I think I meant to ask whether I should try to find an expression for the x-axis. But now I understand that I need only the expression that I got, and integrate it within the limits that I have...in this case, 2 and 0

kc_kennylau · Answer

This is a definite integral, and the limits are the x-intercepts. Do you remember how to find x-intercepts? :)

anonymous · Answer

I think I meant to ask whether I should try to find an expression for the x-axis. But now I understand that I need only the expression that I got, and integrate it within the limits that I have...in this case, 2 and 0$$\int\limits_{0}^{2}$$

kc_kennylau · Answer

yep

anonymous · Answer

then I sub the x's and subtract F(b) -F(a), right?

kc_kennylau · Answer

so that makes $\Large\int_0^2\normalsize (x^2-2x) dx$

kc_kennylau · Answer

yep, except $\int(x^2-2x)dx\ne2x-2$ (That's integration not differentiation!!!)

anonymous · Answer

hmm....then I got lost...again.

anonymous · Answer

I need to differentiate?

kc_kennylau · Answer

no you need to integrate but you differentiated

kc_kennylau · Answer

$$\large\int x^ndx=\frac{x^{n+1}}{n+1}$$

kc_kennylau · Answer

Don't give up :)

anonymous · Answer

$$\frac{ x^3 }{ 3 } -x^{2}$$ ??

kc_kennylau · Answer

yep sorry for late reply

anonymous · Answer

It's quite allright. Thanks for fan-ning. And for your help. =)

kc_kennylau · Answer

but you still have to plug in x :)

anonymous · Answer

Yeah, but these are the details. I now repeated the solving algorithm of these kinds of questions.

kc_kennylau · Answer

good

anonymous · Answer

answer: -1.3(3)

kc_kennylau · Answer

yep