#how do i even do this
87 messages · Page 1 of 1 (latest)
because it wants the area of a curve
and we've been given an equation
so it must be integration
now for integration, we need to find the limits
these are essentially the roots to the equation
so find the roots
(in this q, a is a constant)
find the roots of the equation y=a²-x²
what do you think
use a bit of common sense
we're finding the roots
how would you find the roots of y=x²-5x+6
I mean like
yeah but
what do you have to do
there's a y
how can you find the roots with a y in the equation
set y=0
this is gcse knowledge
about solving quadratics
anyways
?
yeah it is y=0
I was just trying to get you to work it out
anyways
so, find the roots of y=a²-x²
keep going
solve for x
difference of two squares
yep
there are two solutions
yep
so now the integral between a and -a of a²-x² = 36
have you been taught what integration is
the integral is the big s shaped thing
yep so find the integral between a and -a of that equation
read the question
it's not the y
I'm saying the integral between those two limits is = 36
no
y is not 36
the integral
is = to 36
very very different
????
integrate!!!
just integrate it
🤗
for the 4th time
it is NOT equal to 36
the integral is = 36
very different
yes
integrate that
a is a constant
not a variable
we are integrating with respect to x
constants are terms that do not change, like the number 7
variables can change depending on what we want to make them
x is a variable because I could make x whatever I want
a is a constant however because it has a set value
because a² does not integrate to 1/3a³
whenever you are integrating a constant term, you add an x
so integral of a² -> a²x
it's not equal to y
it's just the integral
the y disappears
good
now because we're doing definite integration
the + c doesn't matter
no it's not unless you did your limits with the + c
then yes it is an issue
yes
carry on
yes
no
limits are the things that go on the top and bottom of an integral
they tell us between what values we are integrating
so we can find the area of the curve between those two values
ydp
yep
yes
well done