hi everyone today we're going to talk about how to evaluate the integral of an even function
to complete this problem we'll confirm that our function is even and then simplify and
evaluate the integral let's take a look in this particular problem we've been asked to
evaluate the integral of x to the six plus one on the range negative two to two now the
first thing we should notice with this problem is that we have the definite integral range
here negative two to positive two which means that we're being asked to evaluate a symmetric
area about the y axis so we have here the x axis and the y axis and we want to evaluate
from negative two here to positive two so the same distance from negative two to the
y axis as it is from the y axis to positive two so thats always our first give away when
it comes to simplifying this as an even function the next thing we wan to look for is to test
whether or not our function actually is even if our function is in fact even and we're
being asked to evaluate the same distance to the left of the y axis as we are to the
right of the y axis then we can take advantage of a specific kind of simplification when
it comes to evaluating the definite integrals of even functions so we want to test to see
whether or not our function is even so first of all let's call our function f of x so we
just take everything that's inside our integral here and say f of x is equal to x to the sixth
plus one now in order to test whether or not the function is even we want and go ahead
and plug in negative x for x into our function so everywhere we see x in our function we'll
plug in negative x so we'll get negative x to the sixth plus one and when we simplify
this we'll see that we'll actually get x the sixth plus one whenever we raise negative
x to an even power we're going to end up with a positive sign here and just x to the sixth
so what we see is that we actually end up with the exact same function as we did when
we started with our original function so because these two are the same the original function
and the function we got after plugging in negative x because those two are the same
that means that f of x is even so we can go ahead and say f of x is an even function so
now that we've proven that f of x is even we can go ahead and evaluate the integral
and take advantage of the simplification that we were talking about so because the function
is even and because we're evaluating from negative two to positive two the same distance
on the left of the y axis as on the right of the y axis instead of evaluating from negative
two positive two we can go ahead and evaluate from zero to two and just multiply by two
and the reason is because since we know that the function is even that means that we know
that the function is symmetrical about the y axis and if it's symmetric about the y axis
that means we're going to have the same area in this range as we are in this range and
so we can just take the area on this range here and multiply it by two and it's the same
thing as taking the area from negative two to positive two so we'll simplify the integral
and call it zero to two multiplied by two of x to the sixth plus one now we can just
go ahead and integrate so we'll get one seventh using power rule here on seventh x to the
seventh plus x will be our integral and we're going to evaluate that on the range zero to
two remember that with definite integrals we always want to plug in the top number first
and then subtract whatever we get when we plug in the bottom number so let's go ahead
and plug two in so we'll get one seventh times two to the seventh power plus two minus two
times one seventh times zero the seventh plus zero and obviously we can see here this entire
second term will go away because we're just going to have zero and then the first term
here we'll have two times one seventh times two the seventh which is one twenty eighth
plus two simplifying further we'll get two times one twenty eight over seven plus and
we can find a common denominator here and call two fourteen over seven that gives us
two times one forty two over seven which of course just simplifies to two eighty four
all over seven and that's it that is the value of the integral of x to the sixth plus from
negative two to positive two so I hope you found that video helpful if you did like this
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