#Integrals (3)
134 messages · Page 1 of 1 (latest)
this is what I tried already but aint able to get the right answer (34/3)
Ion get where I am wrong tbf
<@&286206848099549185>
,w x^2+2x-3=0
,w integrate x^2+2x-3 from -2 to 1
,w integrate x^2+2x-3 from 1 to 2
this should be -9
and this should be 7/3
also an abuse of notation here, but I know that we're all a bit sloppy on the computer
@gentle epoch
but this doesnt get me to 34/3 I think
Or either I'm wrong on the final resolution
bcs 34/3 is 11.33333
But I have -9 and 2.33333
the -9 represents the area below the axis from -2 to 1
so the total area is just 9
$9+\frac{7}{3}=\frac{34}{3}$
or since it's the air I get to -34/3 but i have to be pos. ?
Civil Service Pigeon
you're right
thanks
You might have an idea for the 2nd one or you're busy ?
I can try something and send it here
it seems like the same idea
so use the same logic you did
try to avoid computational slips this time tho
yea sure
Ion know where I lost some stuff in my resolution tho, has to be a calculation error
,w integrate x^2+x from -1 to -1/2
,w integrate x^2+x from -1/2 to 2
why are you integrating the antiderivative
$\int^{2}{-1/2} (2x+1) \dd{x}=[x^2+x]^{2}{-1/2}$
Civil Service Pigeon
I need to get the right f(x) nah?
??
What is "the right f(x)" supposed to mean
I need to get to x^2+x
right ?
So I intergrated it
Or either i didn't understood ur question
might be my fault my main language's french so im trying my best
Tu as déjà intégré (2x+1) à (x^2+x)
(Le français n'est pas ma langue maternelle, mais j'ai essayé) 🤷♂️
Sure no worries we can keep going in english tho i u prefer, but wym by am I integrating the antiderivative, like wheres the mistake
so what should i do ?
this ? ,w integrate 2x + 1 from -1 to 2
,w integrate 2x+1 from -1 to 1/2
yes...
Oh sorry then
np
,w integrate 2x+1 from -1 to -1/2
,w integrate 2x+1 from -1/2 to 2
and on the first one, did I had to integrate as I did or it had to stay like this
$\int^{1}{-2} (x^2+2x-3) \dd{x}=\left[\frac{1}{3} x^3+x^2-3x \right]^{1}{-2}$
Civil Service Pigeon
so this was right fine, just a calculator mistake
this is fine (barring the arithmetic mistakes)
this had the same issue I mentioned earlier
alright thanks, if u got time i just got two more, one that looks like these and one for the volumes
can you send a screenshot please
yea sure
ion why they sent like that
so the last one
I need to find the air between the two functions
on this one ion really know how should i do, should i do f(x)-g(x) ?
and from 0 to 1
,w graph y=x+x^2, y=3x
they intersect at 2, not 1
but yeah, just integrate the difference
the line is above the curve so it should be $$\int^{2}{0} (3x) \dd{x}-\int^{2}{0} (x+x^2) \dd{x}$$
Civil Service Pigeon
don't forget the dx either
perfect
lm try it and I'll see if i don't do any calculating mistake
sounds good
this the last one that I got
The idea is that when you rotate the curve, the solid you get has a bunch of circles as cross sections
I’ll let you look up a visual of that yourself
But the distance from the x axis to the curve is your radius
lm translate this bcs I'm not sure
It’s a function of x, so you integrate over the values of x
Also, if that were the case, your upper bound would be wrong then
so if I have somethinf like that at my exam I'll have to go from 0 to 2
Sure
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