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Solve the given problems by integration.Find the root-mean-square current in a circuit from $t=0$ s to $t=0.50 \mathrm{s}$ if $i=i_{0} \sin t \sqrt{\cos t}$

Calculus 1 / AB

Chapter 28

Methods of Integration

Section 5

Other Trigonometric Forms

Integrals

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hi there. And this problem were asked to compute this definite integral. Now it's the definite integral of vector valued function. So from the end of this chapter, let's remember our definition of the integral of a vector valued function, uh is we're gonna have three separate into girls. We can just add together. So it will take the definite integral of this first component which, let's rewrite that as Tito the 1/3 power. As long as we're at it, that will be our I component. Plus again definite integral from 0 to 1 of our function, one over t plus one that were J component and finally definitely a girl from 0 to 1 of e to the minus t. That'll be RK component. So we just to each of these separately, let's begin with this 1st 1 here. Let's find an anti derivative. In order to use the fundamental theorem of calculus of the anti derivative of tea to the 1/3 we bumped power up by one to get 4/3 and then divide by that in front. So we have 3/4 t of the 4/3 we'll plug in one and zero, and that will be our I coordinate next for the J Cordant uh, anti derivative one over T plus one. Now, since it's one over a function, we suspect the natural log of that bottom function. And there's no need to use U substitution here, since a derivative of T plus one is just one. So the anti derivative truly is a natural log of t plus one on again. We will plug in one end zero Jay here and finally for K component this last one here, anti derivative eat of the minus t is minus you. The minus T. We'll plug in zero on one. That will be our key component. Okay, so let's compute these now. So if you plug in 11 of the 4/3 power is just one. So plugging in one gives us just 3/4 times one, which is one plugging zero clearly gives us zero. So that's quick for the I component. Now let's move on to the J component. Ah, natural log of one plus one says natural. Log of to minus. I was plugging zero for tea. Natural log of zero plus one. This natural log of one, the natural log of one is zero on any log of one is just zero. So that's our K component. Is there a J component now for K? I was plugging one for so minus eat of the minus one and minus. That's a minus minus, so we end up with plus and eat. Of the zero is just one. There's RK component. So we're almost done. Just we'll clean this up a little bit. The end up with 3/4 I That's it for the I component plus natural lot of to Jay, that's RJ component. And finally for K. Let's see, we can rewrite this a little queen. Or maybe that's right. The one first and then eat of the minus one is the same as one over E. So that's RK component and we are done

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