# Homework Help: Space between these functions of x

1. Jan 14, 2009

### Dell

i need to find the are trapped between the graphs of
y=-x3
y=arctg$$\sqrt{x}$$
x=2
x=1

y=arctg$$\sqrt{x}$$ will be my top function so the area will be

$$\int$$arctg$$\sqrt{x}$$-(-x3)dx from 1 to 2

in order to to this i can split the integral up and get
$$\int$$arctg$$\sqrt{x}$$dx -$$\int$$-x3dx both 1-2

|$$\int$$-x3dx| (1-2) is 3.75 with a simple table integral

but i dont know how to integrate the arctg

what i think could work is...

arctg$$\sqrt{x}$$=t
tg(t)=$$\sqrt{x}$$

$$\int$$tdx while dt=dx/1+x2

$$\int$$t(1+x4)dt
=$$\int$$t(1+tg4t)dt
=$$\int$$tdt+$$\int$$(tg4t)dt

another option
$$\int$$t(1+x4)dt
=1/2$$\int$$(1+tg4t)dt2
=0.5$$\int$$dt2+0.5$$\int$$(tg4t)dt2

sort of feels like im getting lost in it around here,,, any ideas??

2. Jan 14, 2009

### Dell

lets call t2=p

$$\int$$(tg2p)dp=$$\int$$$$\frac{sin^2p}{cos^2p}$$

=$$\int$$(1/cos2p)dp-$$\int$$dp
=tgp-p
=tg(t2)-t2
=tg[{arctg$$\sqrt{x}$$}2]-(arctg$$\sqrt{x}$$)2
=x-(arctg$$\sqrt{x}$$)2

is this correct

Last edited by a moderator: Jan 15, 2009
3. Jan 14, 2009

### Dick

To integrate arctg(sqrt(x)), set u=sqrt(x). So du=dx/(2*sqrt(x))=dx/(2u). Or 2udu=dx. The integral in terms of u becomes arctg(u)*2udu=arctg(u)*d(u^2). Now integrate by parts.

Last edited: Jan 14, 2009
4. Jan 14, 2009

### Dell

on my table of integrals i dont have an integral for arctg, how do i do this? do i need to do something to the effect of

arctg u =v
tgv=u and then carry on from there (similar to what i posted? ) or is there a abetter way?

5. Jan 14, 2009

### Dick

To integrate arctg(u)*d(u^2) you don't need to know how to integrate arctg(u). You only need to know how to differentiate it. a*db=d(a*b)-b*da. That's how integration by parts works, right?

6. Jan 14, 2009

### Dell

cant see it, could you please show me.

7. Jan 14, 2009

### Dick

The integral of arctg(u)*d(u^2)=arctg(u)*u^2-the integral of u^2*d(arctg(u)). That's what integration by parts means, yes? What d(arctg(u))? Differentiate arctg(u).

8. Jan 14, 2009

### Dell

ive read up a bit on this and found an equation

$$\int$$udv=u*v-$$\int$$vdu

is this what you mean??

so in my case
arctgU is my u function
and u^2 is my v function
where u is the root of x

so

$$\int$$(arctg(u))d(u2)=(arctg(u))*u2-$$\int$$u2d(arctg(u))

now d(arctg(u))=$$\frac{1}{1+u^2}$$*d(u^2)

so $$\int$$u2d(arctg(u))=$$\int$$U2/(1+u2)d(u2)

correct up till here??

=$$\int$$1-$$\frac{1}{1+u^2}$$du2====u2-ln|1+u^2|

which is x-ln|a+x|

9. Jan 14, 2009

### Dick

You are getting there. But d(arctg(u))=du/(1+u^2) (NOT d(u^2)). So the integral you are left with is u^2*du/(1+u^2). You'll find there aren't any logs in the problem.

10. Jan 14, 2009

### Dell

how did you come to d(arctg(u))=du/(1+u^2)
is (in general) d(t) not equal to the differential of t multiplied by dx???

in this case d(arctg(u))= differential of (arctg(u)) * d(u^2)

how do i know which d() to multiply by?

11. Jan 14, 2009

### Dick

Exactly as you said. d(f(x))=f'(x)*dx. d(arctg(x))=(d/dx arctg(x))*dx. d/dx(arctg(x))=1/(1+x^2). So d(arctg(x))=dx/(1+x^2). I don't know where you are getting the d(u^2) stuff.

12. Jan 15, 2009

### Dell

i see my problem, what is the differential of arctg(x^0.5) is it not the 1/(1+x^0.5) multiplied by 1/2(x^0.5) ??

13. Jan 15, 2009

### Gib Z

Well, it appears you are using the chain rule, which is good, but you are forgetting to square something. Look at what the derivative of arctan x is again.

14. Jan 15, 2009

### Dick

No! The differential of arctg(sqrt(x)) is 1/(1+sqrt(x)^2) times 1//(2*sqrt(x)). Like Gib said, you forgot the square in the derivative of arctg. That's EXACTLY the same thing as saying the differential of arctg(u) is 1/(1+u^2) times du where u=sqrt(x). Look at it!