Evalute the integral (what the hell is the alt function for the integral sign, anyway?) from 0 to 1 of (x + (root of) 1 - x^2) by interpreting in terms of areas

What step am I missing?

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Lyinar Ka`Bael had this to say about pies:
Then -1/6 is the answer?

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Answer is still 1/6. When evaluating a definite integral, I'm almost positive you take the absolute value.

GOGO Maple

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Razor stopped beating up furries long enough to write:

GOGO Maple

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That tells nothing about the process of getting there and I'm not going to have Maple to use on a test. Where is it getting that pi?

Lyinar Ka`Bael fucked around with this message on 04-24-2005 at 07:54 PM.

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Lyinar Ka`Bael was naked while typing this:
What about u and du substitution?

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I think that the trig substitution is the only way. I went back and pulled up what the integral answer is w/o the values:

the u sub first then the trig sub second

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We were all impressed when Lyinar Ka`Bael wrote:
What about u and du substitution?

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You cannot use that for this problem. In order to use algebraic substitution you need to be able to put the differential back in the equation. In this case there is no term you could substitute that has its differential in here. Although something like u = 1 - x^2 might be incredibly useful, du = -2xdx which is not in the equation.

The pi comes from the arcsin

Got it.

The answer of the first will be 1/2. It's just a line.

Thanks for the help. I'm sure I'll have more. We sped through these sections.

Lyinar Ka`Bael fucked around with this message on 04-24-2005 at 08:31 PM.

I need to evaluate the integral from 0 to 8 for the absolute value of x^2 - 6x +8

I know you have to get rid of absolute value first. But I can't remember how. Any suggestions?

God I'm stupid. I remember now. Break it separate integrals

Lyinar Ka`Bael fucked around with this message on 04-24-2005 at 08:45 PM.

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The logic train ran off the tracks when Lyinar Ka`Bael said:
Ugh god I hate this class. I hate absolute value, too.

I need to evaluate the integral from 0 to 8 for the absolute value of x^2 - 6x +8

I know you have to get rid of absolute value first. But I can't remember how. Any suggestions?

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You break it up into two functions, one for when the eq is positive, make its domain for all numbers that it is evaluated at to be positive, then another equation that is the opposite of the first, for the domain that the equation would give out negative numbers.

If x < 0 or x > 4 x^2-6x+8

If 2<x<4 -(x^2-6x+8)

That's what I remember at least.

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Noxhil2 had this to say about Captain Planet:
You break it up into two functions, one for when the eq is positive, make its domain for all numbers that it is evaluated at to be positive, then another equation that is the opposite of the first, for the domain that the equation would give out negative numbers.

If x < 0 or x > 4 x^2-6x+8

If 2<x<4 -(x^2-6x+8)

That's what I remember at least.

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When I found the zeros for it, I found it has one segment where it's negative and two where it's positive. This coincided with the graph in Maple.

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This insanity brought to you by Mr. Crabs:
After reading this thread I'm relieved that I've filled the math requirement for my major.

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This stuff isn't too bad. Diff eq is somewhat of a different story, though.

You should see some of the multivariate calculus... the page becomes a total mess.

What's the indefinite integral for 1/x?

-1/x^2 right?

Lyinar Ka`Bael fucked around with this message on 04-24-2005 at 09:30 PM.

Okay, here's a fun one

Find the derivative of the function f(x) = the integral from sqrt(x) to 3x+1 sin(t^4) dt

I thought the derivative of the integral is just the function again?

If it is d/dt then it is just sin((3x+1)^4)-sin(x^2).

If it is d/dt then you have to integrate sin(t^4), which I don't know how to do. Then evaluate it as normal.

Naimah fucked around with this message on 04-25-2005 at 12:17 AM.

If it's calc one, would it be x times root (1-x^2)? Because then you could do an u substitution.

if it's plus, then it has to be trig substitution and I didn't start that until calc 2...

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Naimah got all f'ed up on Angel Dust and wrote:
d/dt or d/dx?

If it is d/dt then it is just sin((3x+1)^4)-sin(x^2).

If it is d/dt then you have to integrate sin(t^4), which I don't know how to do. Then evaluate it as normal.

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It's asking to find the derivative of f(x) and f(x) is the integral between those points of sin(t^4) dt

I took it as taking the derivative of what you get when you evaluate F(B) - F(a) and came up with some weird shit

I really loathe calculus.