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# The graph of a function $f$ is given. Estimate $\displaystyle \int^{10}_0 f(x)\, dx$ using five subintervals with (a) right endpoints, (b) left endpoints, and (c) midpoints.

## (a) 6(b) 4(c) 2

Integrals

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### Video Transcript

Alright, this might be a pretty lengthy explanation. I'll try to do it as quickly as possible. You're dealing with estimating the area using intervals under the curve and using rectangles. So in my example I used the graph one half X squared. And with five sub intervals that means delta X will be 10 0 divided by five. So my change in X will be too, that's going to be the same for yours. What's going to be different or your wife values because I don't know what your function is. So I found my Y value simply by plugging X in. So f of zero equals 0 If I plugged in six for example, one half times six squared is what gives me 18. Now to do the right hand endpoints, I start at the right hand side go up to my graph and then move over. Here's my first rectangle, started the next standpoint, go up to the graph. Move over up to the graph and over up to the graph and over This 1's gonna be little. So I'm finding the area of each of these rectangles and remember area is base times height or change in acts or are distance in between times are Y value. So using the right hand and points, each of these is to so to times. So I'm going to start Going to start with this one. So it's two units across and 50 units high Plus two in its cross and 32 units are high Plus two units across 18 high Plus two units across eight units high Plus students across and two units high. Now you could use a shortcut and you could just do two and then add the Y. Values. But you can see I didn't use this y value I just use these so add those all up and that will give you the approximation using the right hand endpoints. So I paused the video so I could erase those values. And now we're going to do left hand endpoints. Well if I start here there's no height. So there's actually not going to be any rectangle. So the next point that I would use I'd go up to the graph over up to the graph over oops this one actually went a little too high up to the graph it over up to the graph it over. So again all of these have a with of two. So I'm just gonna I'm gonna do this one a little differently. I'm not going to do to every time. So the height for the first rectangle was zero, that's my Y coordinate here. The next one was to eight 18 and 32. And this time I'm not going to use the 50. So add all those together. Multiply. And that gives you the approximation for area using the left hand end point. Now for the midpoint I'm actually gonna need to find the midpoint between each of these and that's the value I'm going to use for the height of my rectangle. So using the mid points this is going to change. And my ex value then will be one because that's halfway between 35 seven and nine. And then I go to my graph and I find the Y. Values at those points one half 4.5, 12.5, 24.5 and 40.5. Again these values are found by taking the X. values and plugging them into my function. So now my graph is actually going to be based on the midpoint here. The height of my graph, The height of my rectangle. Okay, they didn't do that very well. This one's got to go up to about where the here we go and the heights about here. So there's the rectangle for that one. So there's still going to be two units across. Used except now the height of each rectangle is going to be based on that particular midpoint value. So the integration or the area will be approximated by again, still two units across. But now my y values will be .54.5, 12.5 24.5 and 40.5

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Integrals

Integration

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