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Find the volume of the described solid $ S $.

A pyramid with height $ h $ and rectangular base with dimensions $ b $ and $ 2b $.

$V=\frac{2}{3} b^{2} h$

Applications of Integration

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point. So form formula is why is m which is our slope Times X minus X one. Plus why one which means we have why is be over to h Times X. Okay, now that we have this, we know this is the is half the length of what we need, Which means we need to double this to get SFX. So if we doubled us, we know the two just cancels. That's the only thing that really changes. So now we have be over age times X again, No tuks. We're doubling it, which means the two cancels. Okay, now that we have this, we know that we have to multiply the first s by the second s. So before we do that, we know if the first s is be over, H times acts, then for the second part, we have to double it. We have to double this value to the second part, which means that we have simply to be over age times X. So these are US one and acid too. Which means our volume is the interval from zero h of be over H of X. That was our 1st 1 times to be over h of X. That was our 2nd 1 D x. Okay, cool. So now that we've got this, let's pull out the constant. Anything that isn't an X, we can be pulled out. This makes the law easier integrate because we know we can leave the constant alone. That's one of the That's a large part of the term we have here. And we know we can use the power rule, which means increased explain it by one divide by the new ex tenant to get 1/3 X cubed from X squared. Okay, now that we have this, it's time to pull again. We pulled out the 30 pull out the 1/3 which just becomes a three on the bottom because it get Constance can be pulled out. You should have learned this in a previous chapter, which gives us our final answer of 2/3 B squared H