Good day stu dents will come two-part one of the fundamental theorem of calculus part
one in this installment of very going over three examples to get started let's write
down what the fundamental theorem of calculus part one is FTC part one tells us that that
of if F of function F is continuous on the closed interval a be then of the anti-derivative
written by F of X is equal to the integral from a some eight to ask of F of T DT okay
and also the derivative of the anti-derivative F of X is given by the derivative of the anti-derivative
expressed in the integral form which is the integral from a to X of F of T DT is equal
to F of X okay this circle formula just illustrates the inverse relationship between the anti-derivative
and the derivative safety differentiate the anti-derivative you end up with the original
function that you started with okay now let's take a look at the following the instructions
are for us to find find the why the X for number one is you have a function why given
as an integral from three to X square of sine T dt okay right let's take a look at his function
if examining the FTC part one we noted that the argument it into grand is F of F of X
or in this case of F of T okay so in this sum situation right here F of T is equal to
sign T okay now is go ahead and of find the anti-derivative of this so with the have why
equals the expression for the anti-derivative of f is F of of F of T and our going to be
making use of FTC part two of three what that is the second so that of F of T evaluated
from three to X square okay remember what FTC part two it so let's write it down on
the side FTC part two tells is that into the definite integral from a to be of F of X the
X is equal to the anti-derivative of f which is F evaluated at B minus anti-derivative
of F f which is F evaluated at eight so this differences the equivalent to that okay this
is FTC to seven or using FTC one into for these problems right so starting up like this
using FTC part two what would those will plug in the upper limits of integration and a minus
the lower limits of integration so in this this is going to become of the anti-derivative
of F evaluated at X square minus the anti-derivative of F evaluated at three okay so this is one
why looks like using FTC part two now what is what we asked to find the us to find the
why the X okay so is find the why the X to do that we just simply differentiate this
of function the have here so the why the X is equal to be the X of this expression upper
case anti-derivative of F of X square minus anti-derivative of F at three okay so this
is going to become now we differentiate the anti-derivative you end up with the original
function you notice here DDX of F of X F of X is F of X so this is become F of X square
now we have to using chain role here so the input is a function derivative of X square
is to X okay minus the derivative of F of X the derivative of the anti-derivative is
a function so's work is F of three times the derivative of three which is zero okay now
let's go ahead and simplify this this becomes two X now what is F of X square is F of T
is sine T F of X square will simply be what's it to be signed of X square right so you simply
replaces T which is inputs with X square and have that's and then of this minus zero you
final answer is simply to X sine X square okay now regulation initially shortcut you
have you have to do all this work to be the final answer this is why show you how it works
we going to use the shortcut anytime you have a variable in a constant to find the derivative
oh anti-derivative expression like this or you simply do is you inputs the variable a
limits given that the upper limit to the variable is just input that into the function in the
#of X square the have here multiply that by the derivative of that variable limits the
input says which is to X in the goes your final answer the constants PCO bother with
that's because we differentiate it using the chain role the becomes zero in that portion
okay so if I wanted to do this real quick without the all this work is take X square
plugging here that using sign X square and I do not forget the chain role multiply this
by the derivative of the of that's which is to X and there goes your final answer can
do it in less than a second okay right let's Psalm take a look at another example question
number two so we have in this case why equals the integral from cosine X to four of the
square root of one minus C square minus T-square DT okay and what is the shortcuts are in this
case the variable is downstairs so you can end up with negative results and secant have
zero minus this expression so you can you see what the answer is the answer is simply
going to be the square root of one minus cosine square X times negative sign X so the derivative
of of sine X negative sign X the derivative of cosine X is negative sign X right in the
have the minus in front so you simply have sign X times the squared of one minus cosine
square X W final answer all initially the steps is go through all the steps okay in
this situation the integrand is F of T right F of T in this case is the square root of
one minus T square okay now so why equals the integrand the integral from cosine X to
four of the square root of one minus T-square DT using FTC part one of this is going to
be of F of T okay of elegant is FTC part two to evaluate this from cosine X all the way
to for right so FTC part two use requires is to plug in the upper limits of integration
minus the lower limits right so we have F of F the anti-derivative of F at for minus
the anti-derivative of F at cosine X okay now this is why we are asked to find the why
the X so how the find the why the X we just simply differentiates what the have here so
that we looking for DDX of why why is the anti-derivative of X of which is for F of
four minus F of cosine X right that again cosine X
okay right so we going to argue summertime differentiation to evaluate this the first
piece the derivative of F of the anti-derivative of F of four is going to be of F of four times
the derivative of for which is zero minus umm the derivative of F of f of cosine X times
the derivative of cosine X to not forget the chain role please the derivative of cosine
X is negative sign X okay clearly see our answer here is going to be of a do the steps
so you see everything negative sign X now what is negative F of cosine X this is F of
T right here whatever the input is you take a square roots of one minus the input square
is the square root of one minus cosine square X okay now we had a minus there before this
minus is a that here city simplify our final answer is going to be signed X times the square
roots of one minus cosine square X okay and then can even simplify this further on using
identities goes know that sine square X is cosine square X equals one so one minus cosine
square X is equal to sign square X so this is equal to of sine X times the square root
of sine square X which equals sine X times sign X in your final answer will simply the
sine square X okay so that's that. Right let's take a look at question number three okay
so question three we're what we are to find F prime of X but find F prime of X where F
prime of X is equal to the integral from X square to sine X of the square root of T divided
by one plus to be to the T DT okay in this situation Palm a to look is going to be a
look at longer because we do not have any zero piece because we both we have variables
in both sides so we use the chain role when not going to have zero because these are not
constants okay right so of this one we just involved us plugging in the upper limits time
this derivative minus plugging in the lower limits times the derivative and that will
be the final answer right to is go ahead and do it in this situation the integrand is f
of T which is the square root of T divided by one plus to be to the T okay now you look
for F looking for F prime of T which is basically the derivative so the BX of F of X the derivative
of the anti-derivative is D of let some let's apply the fundamental theorem of calculus
part two here so DDX of what F of X is F of X which is the integral from X square to sign
X routine divided by one plus to be to the T the T psychedelia first apply umm FTC part
one I'm going to use FTC part two to evaluate this integral component right so let's evaluate
that first so the integral of this is basically F of F of X again the integral from X square
to sign X of the squared of T divided by one plus to be to the TDT is going to be the anti-derivative
of f which is F evaluated at the upper limits sign X minus the anti-derivative evaluated
at the lower limits X square okay so this is the how to write this integral using FTC
part two okay so the differentiate this is go ahead and differentiate that's so we going
be looking so of F prime of X is in that notation is going to be bracket F of sine X minus F
of X square now these turmeric on differentiation this one if we differentiate anti-derivative
will end up with the lower case function okay using FTC part one the derivative of the anti-derivative
is a function to have lower case F of sine X using chain role the derivative of sine
X is cosine X minus in this case if we find the anti-derivative of F that gives us f of
X square apply the chain role times two X okay now let's figure out what this integral
we just 's the written as the country spiritual to come using here if you switch to the left
is left and exhibits the with him functions are but before we go on the note is that this
is different from the other one because since we do not have any constant we have both parts
of the anti-derivative present okay so F of sine X is what you get when you plug-in sign
into T in our F of T function here so the have the square root of sine X over one plus
to the third is sign X times the derivative which is cosine X to the right here that's
for this portion the upper limits of of integration minus F of X square is evaluating this function
at X square so is going to be the square root of X square divided by one plus to be to the
what's to the X square right to forget the chain role of the differentiate X we have
to X right there okay right is negative pretty to do that we can write this as cosine X cosine
X of times the square root of sine X divided by
one plus to be to the sign X and then minus now the squared of express X and the multiply
that by two X we have two X square divided by one plus to be to the X square right it
can come identity one of so this is basically what the value of F prime of X is right so
that's that the thanks so much for taking the time to watch this presentation because
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