Properties of Integrals and Evaluating Definite Integrals
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Properties of Integrals and Evaluating Definite Integrals


Professor Dave here, let’s take some integrals. Hopefully we now understand the relationship
between differentiation and integration. These are inverse operations, in the sense
that integration requires taking the antiderivative of a function. For basic functions, this is easy to do, but
it gets extremely complicated as the functions get more complex, and we will have to learn
a number of different strategies to tackle the tougher ones. But that will come later. For now, let’s just go over a few properties
of integrals, and then try some simple examples. As we said, it will be a good idea to quickly highlight some important properties of definite integrals. First, if we have some integral of f of x
dx, over the interval from a to b, then if we switch the limits of integration, integrating
instead from b to a, this will be the same as the first integral but negative. This is because we are integrating in the
opposite direction. Next, if both of the limits of integration
are the same number, the integral will be equal to zero, no matter what the function is. This is because we are essentially asking
about the area under a single point, which would be like finding the area of one infinitely
thin rectangle, which is just a line. A line has no area, so this integral equals zero. Similarly, if we have two integrals of the
same function over adjacent intervals, one from a to b and the other from b to c, the
sum of these integrals is equal to the integral of the function over the whole interval, from a to c. This is the same as saying that the area of
this section plus the area of this section is equal to the total area under the curve. So that covers some properties regarding the
limits of integration. Next, when integrating a constant, we will
simply get that constant times the quantity of the upper limit minus the lower limit. So the integral of three from one to five,
will be three times the quantity five minus one, or twelve. This makes sense if you think of the function
y equals three. If we chop this off at one and five, this
is now just a rectangle with a constant height of three, so multiplying by the difference
of these limits is the same as calculating base times height, which gives us the area
of the rectangle. Similarly, if we are taking the integral of
a constant times some function, we can just pull the constant out of the integral, and
we get the constant times the integral of the function. Then, we have the property that says that
the integral of a sum of functions over some interval is equal to the sum of their integrals
over the same interval. So the integral of f of x plus g of x dx equals
the integral of f of x dx plus the integral of g of x dx. The same goes for the difference of functions,
we just change these plus signs to minus signs. Let’s also quickly note that if the interval
of a function that is being integrated is above the x-axis, its integral will be positive. If instead it is below the x-axis, it will
be negative, as these are negative y values. If it crosses the x-axis, with some portion
above and some portion below, the integral will be the net area, or the area above the
axis minus the area below it. Keeping these properties in mind, let’s
move forward and practice evaluating definite integrals. To start out, we will just integrate simple
polynomials. This involves taking the antiderivative of
a function, as we learned in the previous tutorial. Let’s try something like x squared plus
one, dx, from zero to two. First things first, let’s get the antiderivative
of this function. X squared will become x cubed over three,
and one will become x over one, or simply x. Then we evaluate the antiderivative at the
limits of integration, so let’s write F of two minus F of zero. F of two will be one third times two cubed, plus two. Two cubed is eight, so we have eight thirds,
and let’s turn two into six thirds, so that we can combine to get fourteen thirds. Then for zero, everything just becomes zero,
and we end up simply subtracting zero, so the answer is fourteen thirds. Not too bad, right? Let’s do a couple more. How about two x squared plus root x, from
one to four. In finding the antiderivative, x squared becomes
x cubed over three, so let’s just put the three out here to get two thirds x cubed. Now for root x, it may seem like we can’t
evaluate this, but remember, root x is just x to the one half power. So we can absolutely do exactly the same thing,
we just end up with fractions. We add one to the exponent to get x to the
three halves, and then we divide by three halves. Dividing by three halves is the same as multiplying
by two thirds, so we get two thirds x to the three halves. Now we are ready to plug in some numbers. Four cubed is sixty four, times two thirds
is one hundred twenty eight thirds. Four to the three halves power can be evaluated
by first taking the square root to get two, and then cubing to get eight. Times two thirds is sixteen thirds, so we
add these together to get one hundred forty four thirds. Evaluating for one will be a little easier,
since we just end up with two thirds plus two thirds. That’s four thirds to be subtracted from
one hundred forty four thirds, leaving us with one hundred and forty thirds. How about just one more. Let’s try the quantity (x to the fifth plus
two x cubed plus one) all over x squared, evaluated from one to two. This seems much trickier, because it’s a
fraction, but we can very easily express this as a regular polynomial, we just have to divide
each term in the numerator by x squared. For x to the fifth, that gives us x cubed. For two x cubed, that gives us two x. And for one, that gives us one over x squared,
but let’s remember our rules of exponents and instead express this as x to the negative
two. That will allow us to more easily find the
antiderivative. Now we are ready to do what we normally do. X cubed becomes x to the fourth over four,
two x becomes two x squared over two, or simply x squared, and then for x to the negative
two, remember we are still just adding one to the exponent to get x to the negative one,
and then we divide by that exponent, so we put this over negative one. So that means we subtract x to the negative
one, which is really one over x. So that’s our antiderivative. Now we are ready to plug in some numbers. So this will be our final answer. Hopefully after computing some simple integrals,
we can see the immense power of this method. Prior to the fundamental theorem of calculus,
finding areas and distances associated with curvature was incredibly complex, and only
brilliant mathematicians could figure out how to do it. Now with this simple algorithm of finding
the antiderivative, anyone can do it. Now don’t get too cocky, integration will
get much harder than this, but before moving forward, let’s check comprehension.

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