Good day students in this clip were going to be going over some examples on how to find
definite integral's using anti-derivatives before get started with the examples let's
take a look at the formula going to be using the good to be using FTC part two the fundamental
theorem of calculus part two of so if you taken the anti-derivative of an integrable
function from a to b of F of X the X this is simply the anti-derivative of the function
evaluated at the upper limits of integration minus the anti-derivative of the function
evaluated at the lower limits of integration okay so that's basically the formula that
we are going to be using to let's take a look at example one the task is to evaluate the
following week the following so for the first one which is anti-derivatives
to evaluate the integral from zero two pi over four of secant square X the X okay so
the alright the whole idea of him find the anti-of integral's using anti-derivatives
is is working backwards with your differentiation tables so you ask yourself what function can
I differentiate to get the integrand this is the integral right here the argument of
integration what you integrating is known as the integral answer in this case secant
square ask is you integrand to that's your primary focus so the question is you know
of any function that you can differentiate to get of secant square X if you are remember
your differentiation tables you remember that the derivative ofx is equal to secant square
X okay so the derivative of tan X is sick equal secant square X does this tell me tells
me that the integral of secant square X the X is equal to ten X + constants because if
I differentiate any constant of it zero in the derivative of tan X is secant square X
okay so using this idea I can out evaluate find the anti-derivative of secant square
X okay so in this case in this problem the anti-derivative of secant square X is going
to be signed X so this is how right hand X evaluated from zero two pile before I can
use only one bracket here because I don't have any of coefficient in front that I have
the coefficient values to bracket side okay so you can see this problem that the anti-derivative
of secant X the upper case F of X in this situation is ten X right so that we evaluate
team tan X at part of point zero so this is another become tan of the upper limit I will
for minus tan of the lower limits which is zero okay when an attempt I will four is one
minus tan of zero zero you final answer is one okay let's take a look at another example
example number two and what is the definite integral from one to be square of one over
X the X right so let's find the derivative at integral by using anti-derivatives so let's
focus our attention on the integral and what is the integral and in this problem be integrand
is one over X so the remember the function that results to one over X up on differentiation
it is the natural logarithm function call DDX of the natural of them of X is equal to
one over X so this follows that the integral of one of her X the X is the natural of them
of X plus some constants see to the anti-derivative of one over X is a natural of them of X so
can see that in this case F of X is ln X so that evaluate this function at the upper limit
minus the function evaluated that the lower limits using FTC part two okay so let's go
ahead and do that this becomes a natural of them of X evaluated from one to e square we
substitute the upper limits we have the natural logarithm of the square minus the lower limits
LN of one now using up properties of logarithms we can know that these two our counsel out
the inverse is or you can power down this exponent which of the we want to do it I you
going to have of two minus the natural of them of one is equal to zero okay so to minus
zero is equal to two okay and that's the anti-derivative of one of X from one to e square right let's
take a look at another example power of a file for number three there to find the definite
integral from zero two roots three over to of negative one over the square root of one
minus X square the X okay so now of the can clearly see that into grand is negative one
over was one minus X square so that I think about on differentiation rules what function
results in this output up on differentiation this is the derivative of your inverse cosine
function we are cosine function right so the derivative DDX of of inverse cosine of X is
equal to negative one over the square root of one minus X square right to this follows
that the in definite integral of negative one over roots one minus X square the X is
equal to the are cosine of X plus some constants see so in this problem the anti-derivative
F of X of negative one over X square is arc cosine or inverse cosine of X so we're going
to be evaluating this function at the upper and the lower limits of integration and subtracting
the results okay so this becomes the are cosine of which three over to minus the are cosine
of zero okay there cosine of of trick roots three over to this piracy six remember cosine
power six is written to so youre going backwards and they are cosine of zero is pi over 2 to
X of find LCD here to multiply this by three top and bottom find LCD you have of see five
or six minus three pi over six is becomes negative
to pi over six okay to pi over six and you can reduce that's this becomes negative pi
over three and that's the value of your definite integral right us take a look at question
number four so for question four we are to evaluate the definite integral from zero two
three of five X to the fourth dx right so we can clearly see that the into grand here
is five X of the fourth power question is what function what turmeric on differentiation
results and five X of the fourth is one is pretty easy to see recall that the apply the
power rule DDX to X of the fifth power what happens this becomes five X to the four right
so this follows that the definite the in definite integral of five X of the fourth dx is equal
to X of the fifth plus some constants see in this problem the anti-derivative F of X
of five X of the fourth is X to the fifth because the derivative of this results in
the integral right here right so let's go ahead and apply FTC part two is going to evaluate
the anti-derivative at the upper limits so half three to the fifth power minus the anti-derivative
evaluated that the lower limits zero to the fifth power okay all the calculations three
to the fifth power is to hundred and forty-three and that's the value of our definite integral
right stick a look at another example example number five now of what is the in definite
integral from seven two three of three pie the X okay now what's term can you differentiate
to gets three pie as the output other is the constants right so this should be easy to
figure out of the derivative DDX of this constant three pie we in this at the X next to it is
equal to what is equal to three five because we differentiate, variable the first-degree
all you be left with is a constant or the slope right so this follows that, if you in
a great three pie the X this results and three pie X plus some constants see so this follows
that our anti-derivative F of X of three pie is three pie X right so Valley can this of
the upper lower limits of integration we have three pie times the upper limit three minus
three pie times the lower limits seven this becomes nine times minus twenty-one five right
in then of so you final answer it is subtracted tell
is twelve five this is the value of your definite integral's right let's take a look at question
number six ergotamine making an application of can set of integral's definite integral's
to find average value function site so in this problem we are to find the average Value
of the given function underspecified integral for number six the function is sign X in it
on the interval pirate to two three pie over to okay right so before we get started as
the care the formula for the average value of a function real quick is the very basic
formula for the have a value of a function is given by the one over B minus eight times
the integral from a to be of the function dx so that's the formula for the average value
of a function right so the apply to this problem the function under consideration is sign X
as indicated earlier the lower limits of integration are the lower bound of your integral is a
pirate to the upper bound is be three pilot two okay so the average value of sine X is
equal to one over B minus a which is three pilot two minus pirate to times the definite
integral from pirate to all the way to three pirate to of the function sign X the X okay
now find in the value of this of the reciprocal of the right so of the interval right here
is very easy to subtracted we have one over to pie over to you can reciprocate that is
to get him this one cancels out and in this becomes part one over pie right but this is
a different story right here how the find the anti-derivative of sine what we need to
ask ourselves what function results in sign X up on differentiation okay season anti-derivatives
we know that the derivative of of cosine X is equal to what negative sign X right so
we have a situation here we need sign X but this is using negative sine X CR to compensate
for this negativity be of interest is a minus here so from this we can tell that all DDX
of the negative cosine X equals sine X okay so we have to be really careful when the sign
here we doing with trick functions so this follows that the definite integral of sine
X the X is equal to negative cosine X plus some constants see so in this problem the
anti-derivative F of X is equal to negative cosine X so the compute the average value
of this function in this integral we just going to evaluate this function the anti-derivative
at the upper limit minus is value evaluated at the lower limits okay so let's go ahead
and do it on we have the bracket sign of three pie over to I'm sorry ahead of myself the
anti-derivative is cosine so the have negative cosine of three pirate to minus negative cosine
of of pirate to you what's the value of our cosine three pilot today negative and cosine
power to if you Joya cosine curve your remember that does actually the zeros of cosine of
three pirate to and pirate to so you have zeros both their and their so let's see what
we have here this becomes of one over pie times zero minus zero which is equal to zero
okay so the average value of of sine X on the interval I'll pie over to state three
pie over to is equal to zero and this makes sense you think about how the sine curve looks
like it's easy for you to determine on the geometrically that the average value of the
function is actually zero let's let's look at the of graphical significance of that real
quick so let's say we have your tip a cosine function going from here to there to their
only to their and is going to sketch the function on the indicated integral so we know what
the parent sign function looks like so this is pirate to and this right here is three
pirate to have a sign function the is like this and do that again like that from one
its maximum to negative one and what the average the second that's what you think the average
values going to be is getting that in the center right which is the value zero so that
is the average value of the function I occurs that's when X is equal to pie right so that
goes an application of definite integral using anti-derivatives that thanks so much for taking
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