in this video we're going to talk about how to evaluate definite integrals now before we begin take a minute to hit that subscribe button and don't forget to turn on the notification bell so let's talk about the difference between a definite integral and an indefinite integral a definite integral has a lower limit of integration in this case a and b the upper limit of integration an indefinite integral does not have that so this is an indefinite integral and this here is a definite integral the antiderivative of a function f of x is capital f to evaluate the definite integral once you find the antiderivative you need to plug in the limits of integration and so the the value of the definite integral is going to be f of b minus f of a but let me show you the process by which we can evaluate a definite integral so let's start with this example we need to find the antiderivative of each term in this expression what is the antiderivative of 8x cubed well let me give you a review first the antiderivative of a variable raised to a constant is going to equal that variable raised to the constant plus 1 divided by n plus 1. now for indefinite integrals we would have the constant of integration c but when dealing with definite integrals you don't need to worry about c so as an example let's say if we want to find the anti-derivative of x to the fifth all we need to do is add one to the exponent it's going to be x to the sixth and then divide by that result so let's say if we want to determine the antiderivative of 4 x to the 7th so we have a constant times x to the 7th first rewrite the constant and then find the antiderivative of x to the seventh so add one to seven that's eight and then divide by that number and after that you can reduce it so eight is four times two we can cancel the four and so the answer is going to be x to the eight over two so that's how you could find the antiderivative of monomials now let's continue on with this example so to find the anti-derivative of 8x cubed first we're going to rewrite the constant eight then we're going to add one to the exponent three plus one is four and then we're going to divide by 4. now let's repeat this process for the next one so the antiderivative of 3x squared is going to be the constant 3 times x raised to the third power divided by three now what about the anti-derivative of six times x if you don't see a number it's always a one this is six times x to the first power so just like before we're going to rewrite the constant 6 and then the variable add 1 to the exponent 1 plus 1 is 2 and then divide by that result now as was mentioned before because we're dealing with a definite integral we don't need to write the constant c here but we do need to write our limits of integration so now let's simplify this expression 8 divided by four is two so we have two times x to the fourth three divided by three is one so that cancels so we have one x cubed which we can write as just x cubed six divided by two is three so then this is going to be plus three x squared evaluated from two to three now this is going to equal f of three minus f of two and keep in mind this expression here represents lowercase f of x and this expression is the antiderivative which represents capital f of x so we're going to plug in 3 and 2 into capital f of x so in this in these brackets we're going to put f of 3 and here this is going to be f of 2. so let's plug in 3 into this expression so it's going to be 2 times 3 raised to the fourth power plus 3 raised to the third power plus 3 times 3 squared now let's substitute x with two in the second set of brackets so we have two times two to the fourth power plus two to the third plus 3 times 2 squared so this is f of 3 and this here is f of 2. so at this point we just need to do the math three to the fourth power is eighty-one and eighty-one times two is one sixty-two three to the third is twenty-seven three squared is nine times three that's twenty-seven as well two to the fourth power if you multiply four twos you're going to get sixteen and sixteen times two is thirty two two to the third power is eight two squared is four times three that's twelve 27 plus 27 we no longer need the brackets anymore 27 plus 27 is 54. 32 plus 8 is 40 and 40 plus 12 is 52 so this is minus 52 54 minus 52 is 2 162 plus 2 is 164. so this is the value of the definite integral now this value here 164 represents the area under the curve that is between the curve represented by that function and the x-axis between the x-values 2 and 3. so that's what the definite integral can do it can help you calculate the area under the curve but that's the topic for another discussion for those of you who want more examples on evaluating definite integrals check out the description section of this video i'm going to post some links there if you wish to find more examples even harder examples including square roots and other stuff so feel free to take a look at 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