 Hello and welcome to another screencast about fundamental theorem of calculus. This time we're gonna throw in a couple of trig and exponential functions. Okay, so again a brief review here of the fundamental theorem of calculus. So it says that if f is a continuous function on the interval from a to b, and big f is any anti-derivative of little f, then the integral from a to b of f of x dx is equal to big f evaluated at b, minus big f evaluated at a. Okay, so for this example, we've got the integral from 0 to 4 of the function 5e to the t minus sine of t plus 2t to the third power. Okay, so again you'll notice we have differences in sums, okay? So all we can do is just do the integral or the anti-derivative of each piece separately, okay? So that's not a big deal there. This function is certainly continuous. e to the t doesn't have any problems. Sine of t doesn't have any problems, and 2t cubed doesn't have any problems. Okay, so let's go ahead and do the anti-derivative then of each piece. And then we'll figure out what our final answer is gonna be. Okay, so the anti-derivative of 5e to the t. So the 5 is just a constant, so that's gonna be brought along. And then the anti-derivative of e to the t. So again, ask yourself the question, what function has a derivative that's e to the t? Well, that's just e to the t, okay? So the anti-derivative of e to the t is e to the t, and the derivative of e to the t is e to the t, okay? Don't let that one get too confusing on you. All right, we've got a negative sign here. Okay, now we gotta think about what function, when we take the derivative of it, will give us sine of t. Okay, so we know it's gonna be cosine. Hopefully, so then what's the sine, S-I-G-N, plus or minus gonna be on that? Well, remember the derivative of cosine is a negative sign. And here we've got a negative attached to our sine. So technically this is then gonna be plus cosine of t. Okay, so again, if that feels like I pulled a fast one in there, take the derivative of cosine of t and that gives you negative sine of t, which is exactly what's sitting here in the middle. All right, plus, now we gotta do the anti-derivative of 2t to the third, okay? So again, the 2 is a coefficient that just comes along for the ride. We wanna use our power rule then, in reverse. So we're gonna add 1 to 3, that gives us a 4, and then we have to divide by that new exponent, okay? So if we go to take the derivative of this function, the 4 would come down, that would cancel with this 4, and we'd have 2t to the third power left. Okay, we're gonna evaluate this from 0 to, let's see, 4, okay? We could pretty this last one up a little bit, so I'll do that when we go to plug in our numbers. Okay, so plugging in 4, evaluating our function here at 4, we've got 5e to the fourth power plus cosine of 4, plus 1 half 4 to the fourth power. All right, so that's gonna be our first one. So that is our anti-derivative evaluated at our top end point, which is 4, then minus, okay? Then we gotta do the same thing with 0. In some of you, may just try to plug in 0 and think, we're gonna get 0, not necessarily, okay? So if I plug in 0, I'm gonna have 5e to the 0 plus cosine of 0, plus 1 half times 0 to the fourth. Okay, all right, so now let's see if we can figure out what any of these numbers are. 5e to the fourth, well, I don't think that one can be simplified much. Same thing with cosine of 4, but I can at least simplify what this number is. So I believe that gives us 128 when I go to do that. Okay, then minus 5 times e to the 0. Remember, anything to the 0 powers 1, so that's gonna give us 5 plus cosine of 0 is 1. Now, that will be plus 0 at the end, because anything 0 to the fourth power 0 times a half is just gonna give us 0. Okay, so then when I go to combine some of my like terms here, so I have 5e to the fourth plus cosine of 4. And then I'm gonna have 128 minus 6, so I believe that gives me a grand total of 122. Okay, so this would be my exact value for this integral. And if you wanna crunch that out on your calculator, that gave me approximately 394.34. Okay, next one, let's see what other loveliness we have here. Okay, evaluate the integral from negative 1 to 1 of 5 to the theta power plus 2 cosine of theta d theta. Okay, so again, remember this thing here on the back just tells us the variable we're gonna be using, that's the variable we're gonna be integrating in terms of. So all of our functions then should be in terms of theta. Okay, so we know that, like we just talked about e to the t, the anti-derivative was e to the t. Unfortunately though, just like with derivatives, that doesn't quite work the same if you have a base that's not e. Okay, so the anti-derivative of 5 to the theta then is gonna be 5 to the theta. Again, just like the derivative, but instead of multiplying by the natural log of 5, which is what we did with derivatives. We're going to divide by the natural log of 5. Okay, and again, that's because it's the natural log of your base. Fantastic. Then we have plus 2, okay? Cuz again, 2 is a coefficient, so that just gets brought along for the right here. So the anti-derivative of cosine of theta. All right, so we gotta ask ourselves the question, what function when I take the derivative will give me cosine of theta? And that's sine of theta. So that's my anti-derivative. Okay, again, let's check it. Let's do the derivative. So if you do the derivative of 2 sine theta, do you get 2 cosine theta? Absolutely, okay? So that's how you know you've done your answer or you've done your anti-derivative correctly. And now we're gonna evaluate this from negative one to one. And again, we're gonna have some lovely answers here. Okay, so we're gonna have five to the first divided by the natural log of five. Plus two sine of one, okay? So that's one of our numbers. And then minus, we're gonna have five to the negative one, okay? So again, I got this top one by doing my F of B, the top endpoint, which is one. Now I'm gonna go and evaluate my anti-derivative of negative one, my lower endpoint. So that's five to the negative one, ln of five plus two sine of negative one, okay? And unfortunately, I don't think there's much pretty enough we can do with this one. So this would be one way that we could leave our answer. But if you wanna go ahead and approximate that, when I threw that into my calculator, I got 6.35, okay? And again, I also checked to make sure my function was continuous. And yes, I don't have any problems with five to the theta, or I don't have any problems with two cosine theta, so we are good to go. Thank you very much for watching.