use-theorem-55-5-to-evaluate-the-given-integrals-beginarraylllltext-a-int_-112-x-1-d-x-text-b-int_03

Question

Use Theorem 5.5 .5 to evaluate the given integrals.
$$
\begin{array}{llll}{	ext { (a) } \int_{-1}^{1}|2 x-1| d x} & {	ext { (b) } \int_{0}^{3 \pi / 4}|\cos x| d x} & {} & {}\end{array}
$$

Paul Teng · Accepted Answer

always the hardest problem. There are two parts and B. They are dealing with absolute value functions. So we&#x27;re gonna have to split the into gloom, right? Someone function crosses the X axis, uh, or absolute value functions they for give you example of the actual value of X Instead of crossing over the function like that, it would be like this. So we&#x27;re actually need to integrate positive values or need sign. All right, so, uh, who were looking our phone first part? It would be two x minus one. So we need to find when two X minus one equals zero and equals zero at X equals to one half, because when you plug and one happened to two experts when you get zero, and this is indeed between the interval in eight of 1 to 1. So we need to split this function up into two parts. The first would be the integral from native 1 to 1 half of two x minus one DX and second one would be to go from one half to one of two X minus one. The X and you are no longer absolute values, just parentheses. People. Quick fix them Now we have to consider which ones we have to make negative. So the ones that we&#x27;re gonna make negative is going to be the one the stereo, The interval wearing the function is below the X axis. Since that would be normal. However, the absolute value changes for the function to be above changes to swap, to flip over to the positive. Why access? So which one would be normally below the X axis? Well, if you plugged in, made of one into our function to recognize when you get negative two minus one, which is 93. So that is indeed below the X axis. So this would be negative and the other would be positive since it is above the X axis. So evaluating these into Earl&#x27;s uh, um, this all the way evaluating these intervals we get would get negative of your head rivet of two x minus One is going to be X squared minus X on is from made of 11 half founded WAAS Uh, you intended route of up to recognize one&#x27;s again X squared minus X founded by one half to one. So would a plug in our variables. Her ex we get, We get negative times 1/4 minus one half minus. They want to be one minus. Name one plus one. So I will be our first one. Waas, uh, actually minus x 01 minus one or one minus one house. Where the 1/4 minus one half. All right, so if we were to So all this out we get, we&#x27;d end up, we&#x27;d end up with an answer off five house. So that&#x27;s for part A Now, for part B, we&#x27;re dealing with off coast science function, right? So when does the coast and function equal to zero? Uh, well, within the range from 0 to 3. Pirate for those kind of X equals 01 of her ex equals pi over two. Right. We want experts part material. Whole sign of the function equals 20 So now we&#x27;re gonna have to split our function from zero to part of it too. Both sigh, Novaks, dx and then the other, probably from pirate, too. Three pie forms. Of course I d x, of course, that xcx so which one would be negative is gonna be the part of the function that is below the X axis. And we know that co sign starts off above the X axis and then becomes a negative function after pi over two. So it would be a positive for first into go and then negative for the second into girl. All right, so if we were to evaluate, if you were to evaluate this thing to go, we get the integral of coastline of exit. Just the science functions would have sign of acts from zero Zafira too minus the integral clothes. And against will be signed off X from part of a tomb 23 pie Ford&#x27;s. So if you were plugging our variables for X, we get sign up, I over to sign of pi over two minus sign of zero minus sign of three pi over fourths. Um, since we nine minus is that we have made of simple right, very become positive all the plus sign of pi over two. All right, so we know that sign up, however, to is going to be one minus sign of zero is zero. In a sign of three Piper, four seats out of three, power for your baby herded to over to and finally sounded pirate too is going to be one, So our final answer is going to be two minus. It&#x27;s weird to over to.

Paul Teng · Answer

Okay, So for this problem, they want you to solve for the integral of the absolute value of X squared minus one minus 15 over experts, One from negative three to positive three on day. Want you? Uh, this is part eight off to parts. So over here on the right, I have the graph off of the absolute value of expert Nice one minus 15 over extra plus one. And on the interval from native three to positive three. That means we&#x27;re going to be I&#x27;ll draw the boundaries over here, so I&#x27;ll be negative. Three in positive three would be over here on the one thing you can notice is that X equals two X equal to positive to an explosive negative too. At those point, that&#x27;s when the function would cross the X axis whenever, if the absolute value wasn&#x27;t there. All right, but across the x axis. So because of the absolute value sign, Although, although why values have to have apology value, so none of them can be a negative number. So, looking at this graph, you know that the function that I&#x27;m gonna highlight in red, used to parts are normally above the no wonder map normally above the X axis, no matter what, even if you take away the extra absolute value so they will always be above the axe access. However, this middle part right here would be below the X axis if the absolute value signs were not there. So they were reflected when the absolute value signs were added. So you have to keep that in mind with integrating. So when we integrate this part, the middle part, we would have to multiply it by a negative. And another thing we can note is that this is a symmetrical function across the Y axis, right. And we can use the serum. You can use a property of the integral of that. Uh, if they&#x27;re symmetric across the y axis, especially like functions like even functions where we&#x27;re allowed to do seeing the function. And since we&#x27;re multiple, we&#x27;re integrating it from later three to positive three. We&#x27;re allowed to do this or a lot to integrate from halfway. Really, really need to integrate half of it from 0 to 3, and all they do is multiply it by two. All right, this is, uh so make it easier. Since pulling in the value of zero is always well, who would probably be easier than playing in negative three. So now that we have are integral are you just rewritten it. Let&#x27;s go ahead and split the integral but into two parts. So if we were to split it, our first into go will be from will be from zero to 0 to 2. And this will be negative because if we shaded in blue, since it&#x27;s supposed to be below the X axis without the absolute value sign and it&#x27;s X squared minus one minus 15 over X squared plus one the X plus the second goal, which will be to time the integral from to the three off X squared minus one minus minus 15 over X squared plus one DX. And if I wanted to find the integral of these functions, find the anti derivatives of the expressions with respect to acts and if I were you to do that, you know that, uh, we want to do that, we can note that they ended rooting for extra minus one minus 15 over extra pills. One will become excuse over three minus X and not the, uh, the in a derivative of one over expert plus one is eat our candy in function So we get minus 15 arc Tangent O X When this is from zero to then we get plus plus two times was two times are second into roads will be excused or three to over three minus X minus 15 are Qianjin of X This will also be from two or three right? And if I were to plug in these values floors already plugging these values for three running out of space But but we&#x27;re plugging these values for exactly one degree for X, I get an answer off 60 are changes Oh too minus 30 are tangent off three plus 28/3 and this will be your final answer and it looks pretty ugly. So if you approximate it was approximately equal to approximately you book a 38 0.29 All right, so that would be your answer for pro Day now Part B uh, swelling Now Here, they want you to evaluate the absolute value of one over the square it off. When my sex word minus two squared up to from zero to squirt of three over to and the and looking at this, um, your report, the square root of three over to is just a little bit less than one. So I want to draw on my bounds. It would be from zero to approximately three square, to which I&#x27;ll say is somewhere around there. And the one thing you notice is that on the you know that this point right here, this is where the function would cross the x axis. So because you can tell from the sharp turn and of the function and this is when one over discovered of one minus X squared because to explore it too. And that happens when x when accident y when x able students were it to over to. All right, so we know that we&#x27;re going to split our to enter worlds one from zero to skirt of to over to the other from according to or to to wrap 3/2. And now we have to know which ones are gonna which into girls going to be a negative image from will be positive. Uh, well, we know that the area passionate and read the area should be blog X axis However, the absolute value chain uses so that it&#x27;s a boat. The X axis on DSO when we integrating this into go has to be multiplied by negative number and this region here from started to to skirt three or to shay and blue that&#x27;s normally positive above the ex access, so that would just stay as a normal integral. You just remove the absolute values symbol. So Ford and the 40 interval from zero to skirt of two or two that&#x27;s multiplied by negative one because it should be normally below the X axis. So we were right to minus square root of one minus X squared. The eggs plus sent second girl, which is from heard of two over to his word of 3/2, and that one is from his use of the integral of one minus X squared, minus word or two. The X And if I were to ah, I were to find Thean Headroom. It is all these functions with respect to acts. Well, I know that the editorial develops. Word of two is X squared of to and the anti derivative of one over the square root of one minus X squared is going to be our sign. Or you can find the anti derivative off. Negative one over, sort of one minus X squared, which is our close alliance. It will be part of our positive Arcos. I&#x27;ve It doesn&#x27;t really matter here. You&#x27;d get the same answer either way. So I&#x27;ll just go with the negative arc sine route. And the bones are from zero to square it up to over two plus, Yes, uh, second little girl, which is going to consist of arc sine of X minus, X rayed to and it is from zero was not from zero is from started to over 22 We&#x27;re at 3/2. All right, So if I were to plug in, if I were to plug in the values the bound dollies for X, I get get one minus arc sine mark sino swear to over to plus our sign of swearing at three or two minus the square root of six. We&#x27;re tu minus Mark sign Well spirited to over to minus one. And if I were to solve all this out, I get an answer off. Tu minus pi over six. My skirt of six over to when this would be your answer, and this would approximately equal to approximately equal to 0.2 52 if you estimated them.

Paul Teng · Answer

What? All right, So for this problem there there are two parts and be for part 81 unifying the to go on a 1 to 1 of the function absolute value of E to the X minus one over here on the right, I have, um, function absolute value of you to the X minus one. And we&#x27;re integrating from native one to positive one. Right? So mark these boundaries, it was a black vertical line. And, um, because it&#x27;s an exponential function, we can note that when X is look, next is smaller than zero. Uh, these values the area that I&#x27;m highlighting and blue, this is going to be negative since my people Oh, it&#x27;s gonna be below the X axis if the absolute value wasn&#x27;t there. And this green area on highlighting is going to be your positive area because the function is normal. You know the value of eat X 80 x minus one equals two B T X minus one with max is greater than zero. Those would be positive. So if we were to split our to go, we&#x27;re gonna split it down the middle. So integrate from negative one is your own since at zero. This is where the function crosses the x axis. We&#x27;re gonna be integrating, uh, each of the X minus one, but we&#x27;re gonna have to negate it because it is below the X axis. Um, at that at interrupt during that interval, you go and then our second interval would be from 0 to 1 of the normal function et Xbox one, since that is above the integral. All right, So if we were to evaluate these and their gold by fighting their anti derivatives, we get X minus E to the X from a 10 first into grown plus e t x minus x from 0 to 1 for a second. Integral. And then we were not plug in our values. Uh, we were planning our values about your values for X. We end up with negative one plus Heaton a one plus one plus e minus one minus one. And you can I know that, you know, plus one in the minus one. So those cancel out. So we end up with the final answer off E plus e to the negative one minus two. So that is your final answer. For part a so that part being is dealing with the integrating the function, the absolute value of two Money&#x27;s X divided by X from 1 to 4. I&#x27;m gonna mark the boundaries black again. So from one before on and since, Uh, since it crosses since a crosses the X axis at X equals a two, we&#x27;re going to split the integral so that we&#x27;re gonna have to integrate one from absolute value. We have integrate one from one to and then the other. We&#x27;re gonna have to integrate able todo the other working that&#x27;s integrate from 2 to 4. And now I&#x27;m going to decide which one is going to be negative and which ones learning positive and to calculate that we can just plug in a value in between each intervals. And then you get that for the interval from one to, uh, the area on highlighting in red. For this, this is going to be above the X axis, normally 40 valley part a function to my sex over X. So this will be night positive. And for this into pull over here on highlighting and green, this area is usually it&#x27;s going to be below the X axis if the absolute value of them. They&#x27;re so this will be usually negative. All right, So, uh, so if we were to look at a double inter room our first thing to go, we need to keep positive because because it is normally above the x axis to my sex over axes above the X axis. However, the function to minus X over X is not above the X axis during the interval from 2 to 4. So we just have to negate it. So I just swapping and make it X minus two. So we have do now is integrate this functions integrate these double intervals. So if I were to first thing I want to do is to split the numerator and denominator. So we have two over X minus one. The extra first, integral. And you know, from 2 to 4 off, one minus two over X for a second, Integral. All right. So far to calculate the anti drew it is for these expressions. Uh, you get to Ellen of absolute value of x minus x from wanted to waas X minus two. Ellen of absolute value backs from 2 to 4. All right, so if I were to then plug in my boundary points for X I get from I get four Allen of to minus two on a four plus one. And we know that too well informed can be written as to Eleanor two squared, which then becomes for Ellen off to. And then that means that we can that piece cancel out. So you just end up with a final answer off one.

Subhadeepta Sahoo · Answer

and this was in the given into milk and merit tennis. Score in tourism off minus one toe three d x minus. I went. The rest are not minus one. Put three x p x for two whole years. Enough! The ex, his ex minus one minus pi. Here&#x27;s enough Excel Explorer by two. My last one with no record to four times three minus minus one minus five times minus. So I&#x27;m solving this ticket. God said it.

Subhadeepta Sahoo · Answer

mind. Given integral, we conduct this as a case from one Xbox last into symptoms. You go to one who am through broader on my expert MPX disappoint. 20 years enough. XXX right. The one less interesting upto daughter of one man&#x27;s expertise to times next to one minute less one signing most except one from river. Want my one dripping this? You can enter it one less by like Mrs.

Subhadeepta Sahoo · Answer

then this wasn&#x27;t everyone in typical can make better told us in case and some minor toe the one minus three times in two years on From my photo to more text this is a one toe from my toe like this clear things. Now this well connected test on definitional More taxes, it worker xfx that record an zero and minus six. It&#x27;s extracted. And you know, this is not like the rest my not Placido minus x Yes, plus some from you go toe to X dx. Okay, so they should take Marko 19 of the exit X from my knuckle toe to by not three times minus 60 x is minus X squared. Right to, um, my foot, plus increasing up. Expected exploit rightto no little Toto. It&#x27;s a big Michael to Linus. My next to minus three times off is hell. Thus I two squared. What? Do less, Bruce? Well, my knitting Not in the quarto war, minus three times self full. The biggest part of my message. You seem fine. And I&#x27;m sorry

Problem

Use Theorem 5.5 .5 to evaluate the given integral…

Problem 31 Hard Difficulty

Use Theorem 5.5 .5 to evaluate the given integrals.
$$
\begin{array}{llll}{\text { (a) } \int_{-1}^{1}|2 x-1| d x} & {\text { (b) } \int_{0}^{3 \pi / 4}|\cos x| d x} & {} & {}\end{array}
$$

Answer

(a) $\frac{5}{2}$
(b) $2-\frac{\sqrt{2}}{2}$

Related Courses

Calculus Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 5

Video Transcript

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

Integration Review - Intro

Definite vs Indefinite

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

(a) $\frac{5}{2}$(b) $2-\frac{\sqrt{2}}{2}$

Calculus Early Transcendentals

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

Integration Review - Intro

Definite vs Indefinite

Howard Anton, Irl Bivens, Stephen DavisCalculus Early Transcendentals

Grace H.

Catherine R.

Anna Marie V.

Caleb E.

(a) $\frac{5}{2}$
(b) $2-\frac{\sqrt{2}}{2}$

Howard Anton, Irl Bivens, Stephen Davis

Calculus Early Transcendentals