Original subtitles

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Today we're going to be talking about how to evaluate a definite integral.

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And in this particular problem we've been given the interval from negative 1 to 1 of either the negative

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2 x x.

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Now what makes this a definite integral which is visibly obvious is these upper and lower limits of

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integration that are attached to this integral notation here.

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The negative one and the positive one.

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If you don't have those little numbers right there and you just have your integral notation and then

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let's say either the negative 2 x x without these numbers then we call it an indefinite integral where

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there are no upper and lower limits of integration.

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Remember that an integral is all about finding area under the curve.

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So without these limits we're saying find the area under the curve everywhere under the curve and above

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the x axis from negative infinity to positive infinity or from wherever the graph intersects the x axis.

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But in this particular case where we added these limits of integration the negative one and the positive

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one were saying give me the area under this curve and above the x axis between X equals negative 1 and

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X equals ones on that interval.

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That's essentially what we're finding here.

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And definite integrals are just like indefinite integrals in the sense that we're going to evaluate

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the integral but then we have to plug in our upper and lower limits of integration.

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So we'll look at that when we get to it.

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But the first thing we want to do is just evaluate this integral and you'll remember that the integral

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of each of the X is just to the X but when we have this coefficient on the exponent here this negative

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two value we have to divide by that coefficients are integral will be easy to the negative 2 x.

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But then we need to divide by that negative twos will say 1 over negative 2 will be our new coefficient.

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This is opposite of when we take the derivative of E to the negative 2 x we get negative 2 to the negative

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2 x instead of just multiplying by that negative 2 coefficient where dividing by the negative two coefficient

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because we're going in the opposite direction taking the integral.

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So that's our integral and two things to know before we evaluate this integral on the interval negative

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1 to 1.

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First we're not going to add the constant of integration.

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See Remember normally when we take an integral we add to this plus C we're not going to do that because

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when we have a definite integral we don't need to add some arbitrary constant.

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The fact that it's a definite integral removes the ambiguity that would normally exist with an indefinite

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integral where we have to add that constant integration.

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But because we have a definite integral we never have to add that plus c so we won't be adding that.

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And then the way that we didn't know that this is a definite integral for which we're going to evaluate

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on a specific interval is we draw this line here on the right hand side and we say negative 1 to 1.

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So these upper and lower limits of integration here transfer from the integral to this kind of notation

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here and what this tells us is that we're evaluating on the interval negative one positive one.

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So how do we evaluate now.

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Well when you're evaluating a definite integral you always plug in the upper limit of integration first.

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In our case that's one here.

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We plug that in first then we subtract whatever we get when we plug in our lower limit of integration.

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Negative 1.

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So first of all plug in 1 will get 1 over negative.

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To each to the negative two times positive one our upper limit of integration.

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Then we're going to subtract whatever we get when plug our lower limit.

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Keep in mind that sometimes because you're subtracting here you can end up with a double negative so

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it's really important to have parentheses here so that you make sure you distribute this negative sign

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across multiple terms.

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If you have them.

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So we're going to say minus and then 1 over negative to E to the negative two times are lower limit

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of integration here.

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Negative 1.

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So negative 1 like this.

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And now it's just a matter of simplifying whatever we have here.

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So we've got negative 1 1/2 E to the negative to the to the negative 2.

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Now we have minus a negative one half which is going to give us plus one half B to the negative two

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times negative one gives us positive 2 and we could leave our answer like this.

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But it's nice to always lead with a positive term instead of a negative term.

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If we can.

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So we're just going to flip the order of the terms we're going to say one half the squared minus 1 1/2

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E to the negative 2.

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Then if we want to go ahead and factor out a 1 1/2 we can still say 1 1/2 times the squared minus E

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to the negative 2 it would be nice not to have this negative xponent here's what we'll do is we'll move

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that to the denominator which will make the xponent positive instead of negative.

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So we'll get e squared minus 1 over e squared and that negative becomes a positive tool and move it

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to the denominator and that's it that's our final answer.

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That's how you evaluate a definite integral.