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Household electricity is supplied in the form of alternating current that varies from 155 V to -155 V with a frequency of 60 cycles per second (Hz). The voltage is thus given by the equation$$ E(t) = 155 \sin (120 \pi t) $$where $ t $ is the time in seconds. Voltmeters read the $ RMS $ (root-mean-square) voltage, which is the square root of the average value of $ [E(t)]^2 $ over one cycle.(a) Calculate the $ RMS $ voltage of household current.(b) Many electric stoves require an $ RMS $ voltage of 220 V. Find the corresponding amplitude A neededfor the voltage $ E(t) = A \sin (120 \pi t) $.

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Integration Techniques

Missouri State University

Baylor University

University of Nottingham

Idaho State University

Lectures

01:53

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

27:53

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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household electricity is supplied in the form of alternating current that varies from one hundred fifty five bowls to negative one hundred fifty five bowls with the frequency of sixty cycles per second. The voltage is given by the equation. E. Of T is one fifty five sign of one hundred twenty piety. T is time in seconds. Bo meters read armas both edge, which is the square root of the average value of E square over one cycle. So if we have sixty cycles per second, that means one cycle is one sixty years of a second sulfur part eh were given. Aramis, is this square room of the average value of E. T Square. So before we deal with the square root, let's just go ahead and use our formula for the average value to simplify this average value of east clear. So buy a formula for average value. We have won over B minus A. So is just the starting point with zero and B is just one second or so one over sixty, then the integral from A to B and then we have e square. So this is just using the definition of the average value and the fact that one cycle corresponds to one sixteenth of a second. So let's go to the next page. So let's pull out the constants. Here we have sixty times one fifty five square, then the integral Cirone over sixty of science where one twenty. So at this point we can use one of our formulas for Science Square. So the one that's going to be useful here is science. Where t is one minus cosign. Tootie over two. Here we have. So maybe let's use a different different letter than tea, since he's being used already. His ex. So in our problem Exes. One twenty Piety So it's got and use this sixty times one fifty five square. Let's go in and pull off that one half from this formula in a girl. Zero one over sixty one, minus co sign of two times x. So we have two forty piety. Then let's go ahead and integrate this Linda sixty over too. One fifty five square T minus. Sign two forty Piety and the sign is divided by two forty pie in my help you here. So you go ahead and use the U substitution for the general. In the end, points are zero and one over sixty. That's gotta plug goes in so you could go ahead here and divide sixty over two hundred and thirty plugging in one over sixty for team. So two, forty times one over sixty is to forty over sixty, which is for so we have four pie there. So this is for plugging in one over sixty. And then when we plug in zero for tea, we have zero and then sign of zero zero. So we subtract nothing. Zero. So at this point, we could cross those off and then just simplify this expression over here. And if we do so, we should get one fifty five square over two and then therefore the armies. Because remember, we didn't deal with the square root yet from the previous page. So now we have to take the square root. And so we have one fifty five over route too, which is approximately one hundred ten volts. So this is our answer for party. Let's go on to the next page for purple. So for part B were given that the requirement for stoves is that the Aramis has to be two hundred twenty bolts. And we're also given the formula for me here. So Eve t were not given the amplitude. So we have to replace it with some letter, eh? And so, using the definition of the arm s we need to solve the following equation. So two twenty equals armas and there are misses the square root of the average value of the square. And then we need to solve this for a So let's simplify this by square. In both sides, we have two. Twenty square is the average value of he's clear, and then we can go ahead and plug this in. So we have one over, just like in party. The only difference now is that instead of one fifty five, we have to use the letter A for the amplitude four. So, Scott, and simplify this two twenty squared and then hear this one over one over sixty sixty. We could also pull out the A square. And just as in party, we can go ahead and use a protagonist identity here. Let's use the identity. So here we have X is one twenty pi tea. So let's pull out that one half outside the integral. So we have it to here. Integral zero to one over sixty one, minus sign of two ex excuse me co sign. So two times X will be two forty Piety and we can go ahead and integrate this. So let's go to the next page. You knew that we have two twenty squared on the left. On the right. We have thirty eight squared and then the end scroll of one with respect. Tea is tea in an integral of the co side term is signed two forty piety all over two. Forty five and then once again, our end points or zero toe one over sixty. So squatting plug goes in, That's thirty a square, one over sixty minus sign of for piety. So sign up for pie and then when we plug in zero for tea both of the terms of our zero. So there's nothing to subtract And we also know that sign of four pi zero so we could ignore that term. So we have a squared over two. So at this moment, we have on the left side to twenty square. On the right hand side, we have a squared over two so we can solve this for a square. So we have a squared equals two times two twenty square so that a is the square root of this. So if you'd like, we can go ahead and find the exact answer. Radical too. Times two twenty. Or we could approximate this two, three, eleven both, and there's a final answer.

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