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Numerade Educator

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Problem 1 Medium Difficulty

Let $ I = \displaystyle \int_0^4 f(x)\ dx $, where $ f $ is the function whose graph is
shown.
(a) Use the graph to find $ L_2 $, $ R_2 $, and $ M_2 $.
(b) Are these underestimates or overestimates of $ I $?
(c) Use the graph to find $ T_2 $. How does it compare with $ I $?
(d) For any value of n, list the numbers $ L_n $, $ R_n $, $ M_n $, $ T_n $, and $ I $ in increasing order.

Answer

a.$L_{2}=6, \quad M_{2} \approx 9.6 \quad R_{2}=12$
b.$L_{2} :$ under $\quad M_{2} :$ over $\quad R_{2} :$ over
c.$T_{2}=9<I$
d.$L_{n}<T_{n}< I< M_{n}< R_{n}$

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Video Transcript

So we have the I is equal to the integral from 0 to 4 of f of x dx, Um, And for l two, what we'll have is two times eft zero plus two times f of two. And that's because our Delta X is equal to two. R n is equal to two. So we're gonna have intervals from 0 to 2 and 2 to 4. This is going to be two times one half plus two times 2.5. So l two will equal six m two. On the other hand, we're going to have two times f of one plus two times f of three. That's gonna give us two times 1.6 plus two times 3.2, which is going to give us a 9.6 then. Lastly, we have our two, which is equal to two times F two plus two times F four. That's gonna give us two times 2.5 plus two times 3.5, and that's gonna give us 12. So these are our values for l two m two and our tube. So now we go on to part beat here. We see that, um The graph of F is increasing in concave, down the rectangles based on the left interval and points will have space between the tops of the rectangles. So if we look at, say, a curve whenever we have because it's concave down, it's increasing in conclave down. So it's like this when we look at the left hand estimates we're gonna see a bunch of spaces below the curve right here. Space here, space here. So what that tells us is that it's an underestimate. So l two it is under. However, when we look at the right hand, what we end up seeing is that we get something like this. So because of that, you see all this space above the curve, and for that reason, it's going to be overestimate. So those are the slightly easier ones. The harder one is going to be M two, which, um, based on the fact that concave down and increasing the rate of increase is decreasing. So we see that because of that, the rate of increase is decreasing. Its not going like this because that we see that M two is also going to be overestimating for one C um we can use the trapezoidal rule. So what that's gonna look like is Delta X over two times f of X, not plus two times f of x one plus f of x two. And we get that this is going to equal to because that's our Delta X Times two time, 0.5 plus two times 2.5 um, plus 3.5. And that's going to give us approximately nine, which is less than the actual area. But if we were to draw a trapezoid, it's clear that we're getting closer to our value. So we know that nine is an underestimate. So nine is less than I, which is the actual integral. Then for part, Dean, Um, counting squares. We see the area is about 9.1, so we obviously got closer. So based on the reasoning, um, the way that we're going to do it is the left hand estimate is going to be less then the trapezoid method, which is going to be less than the actual integral, which is less than the midpoint, which is less than the right hand every month. Some, Yeah, so this is going to be kind of our inequality to determine what the least is and what the greatest is for. Problem one