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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
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