5

7 Showdx f() fsg' ( + dx g(x) g(x) Tg(x)]2 Hint: start with f(x)T g8. Find| evdx...

Question

7 Showdx f() fsg' ( + dx g(x) g(x) Tg(x)]2 Hint: start with f(x)T g8. Find| evdx

7 Show dx f() fsg' ( + dx g(x) g(x) Tg(x)]2 Hint: start with f(x)T g 8. Find | evdx



Answers

Find an expression for the integral which contains $g$ but no integral sign. $$\int g^{\prime}(x) \sin g(x) d x$$

Given that we have the whole area of the integral from 2 to 8, and that is 7.3. Let's go ahead and just sketch that out here. I just drew an arbitrary graph, so this is not exactly what it looks like. Of course I'm just using this as an example. But that tells us that the area here under the curve would all be 7.3. And then we're told from 2 to 4 that the area is 5.9. So then this area here would be 5.9. The question is, what is the area than from 4 to 8. So this area here, what would that evaluate too? And to do that, we could simply subtract the total area 7.3 minus 7.3 and then minus our 5.9. And that will end up giving us just go ahead and do that out. Six. That was before and 1.4. Therefore, the missing area here must be 1.4. Because if we were to add that area 1.4 plus 5.9, that would then get us back to 7.3 as well. So that is what the missing area for the integral from 4 to 8 would have to be here.

Problem number one give you the X squared except one and do you to go to 16 x.

We want to find the integration to this function. First of all, we recognize that is an improper function, because the power in the numerator is greater than the power in the denominator here. So let's do long division extra for four divided by X minus one. So we have X Cuba little happiness, so we'll get executed. So plus now X square. So we have excuse yet Do we have X square again? So we had plus x x square your ex here. Now remember, it's one. So first we're putting down this fraction into execute plus X square plus X plus one plus one over X minus one. Yes, so excuse when we integrate with X for over four x squared when we integrate the X cubed over three extreme ingredient X squared over two one women integrate. We will have X and one mobile X minus one. When the great we get long Model X minus one and finally plus C

We have G f R is equal to the integral from zero to our of the square root of X squared plus four D X. Now we're going to go ahead and use the fundamental theorem of calculus. Take the derivative with respect to our that's going to be just the in a grand evaluated at our


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