hi everyone today we're going to talk about how use properties of integrals to simplify
a definite integral to complete this problem we'll identify the properties we need use
the properties to simplify the definite integral and then evaluate the definite integral let's
take a look in this particular problem we've been asked to use properties of integrals
to evaluate the definite integral of four plus three x squared from zero to one so in
order to complete this problem we're going to be looking at a couple different properties
of integrals the first one being that if you have an integral that is the sum or difference
of multiple functions you can split them apart into separate integrals so for example we
have two different pieces of our integral here one function which is four and the other
function which is three x squared because they are added together if they we're multiplied
or divided one by the other we wouldn't be able to do this but because they're added
together and if they were subtracted together from one another we could do the same thing
we can split them apart into two separate integrals so the integral from zero to one
of four plus three x squared dx is the same thing as the integral from zero to one of
four dx plus the integral from zero to one of three x squared dx so we just pulled them
apart and again the goal of using properties of integrals is to simplify integrals as much
as possible so that's our first property our second property is that we can pull constants
out in front of our integral so for example whenever everything is multiplied together
inside an integral as we have here right we have three times x squared times dx all three
of these are multiplied together whenever everything multiplied together or divided
we can a constant out in front of the integral so we have a constant coefficient of three
that can come out in front here we have the same thing in our first integral with four
times dx because we have the constant coefficient of four it can come out in front that's to
be distinguished from what we had in our first step here we have four plus three x squared
in this form four plus three x squared we can't pull a four out in front or pull a three
out in front because these two terms are added together only when everything is multiplied
together can we pull out the constant coefficient so to simplify we'll pull four out in front
and we'll get four times the integral from zero to one dx plus pulling our three out
in front three times the integral from zero to one of x squared dx now that's as far as
we can simplify our integrals using the properties of integrals so now we need and go ahead and
evaluate our definite integrals which means we'll be evaluating these on the range zero
to one so to take the integral of dx the integral of dx is just x so we'll get four times x
and we'll be evaluating x on the range zero to one then over here we get plus three the
integral of x squared when we have a power like this x squared we just add one to the
exponent so two plus one gives us three and then we divide the coefficient which right
now happens to be one by the new exponent three so we get one third like that and we're
going to be evaluating one third x cubed one the range zero to one now before we evaluate
either of these on the range zero to one we can just go ahead and simplify as much as
possible so we'll get four x evaluated on zero to one we'll get as you may notice three
and three here to cancel from the numerator and denominator and we'll be left with plus
x cubed from zero to one and here's another property we can use because we're evaluating
both of these terms on the range zero to one four x and x cubed we can do them separately
or we can combine them so we can say actually four x plus x cubed all evaluated from zero
to one and if we want to remember that we're evaluating a whole thing we can put parenthesis
around it like this so now what this tells us to do is when we have a function here and
then we're saying we're evaluating from zero to one we're going to plug the top number
one in first we're going to plug one in for x the four x plus x cubed then we're going
to subtract whatever we get when we plug in zero so we'll plug in one first we'll get
four times one plus one cubed and then we always subtract whatever we get when we plug
in the bottom number in this case zero so four times zero plus zero cubed like this
and now that we've plugged in one and zero we just simplify as much as possible so we'll
get four plus one over here minus zero plus zero and obviously this is all going to go
away and we'll just be left with five and that's it five is the integral of four plus
three x squared on the range zero to one so I hope you found that video helpful if you
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