Articles

5.5E & 5.6E U-Substitution Exercises

5.5E & 5.6E U-Substitution Exercises


We are searching data for your request:

Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.

5.5: Substitution

In the following exercises, find the antiderivative.

261) (displaystyle∫(x+1)^4dx)

Answer:
(displaystylefrac{1}{5}(x+1)^5+C)

262) (displaystyle∫(x−1)^5dx)

263) (displaystyle∫(2x−3)^{−7}dx)

Answer:
(displaystyle−frac{1}{12(3−2x)^6}+C)

264) (displaystyle∫(3x−2)^{−11}dx)

265) (displaystyle∫frac{x}{sqrt{x^2+1}}dx)

Answer:
(displaystylesqrt{x^2+1}+C)

266) (displaystyle∫frac{x}{sqrt{1−x^2}}dx)

267) (displaystyle∫(x−1)(x^2−2x)^3dx)

Answer:
(displaystylefrac{1}{8}(x^2−2x)^4+C)

268) (displaystyle∫(x^2−2x)(x^3−3x^2)^2dx)

269) (displaystyle∫cos^3θdθ(Hint:cos^2θ=1−sin^2θ))

Answer:
(displaystylesinθ−frac{sin^3θ}{3}+C)

270) (displaystyle∫sin^3θdθ(Hint:sin^2θ=1−cos^2θ))

271) (displaystyle∫x(1−x)^{99}dx)

Solution: (displaystylefrac{(1−x)^{101}}{101}−frac{(1−x)^{100}}{100}+C)

272) (displaystyle∫t(1−t^2)^{10}dt)

273) (displaystyle∫(11x−7)^{−3}dx)

Answer:
(displaystyle−frac{1}{22(7−11x^2)}+C)

274) (displaystyle∫(7x−11)^4dx)

275) (displaystyle∫cos^3θsinθdθ)

Answer:
(displaystyle−frac{cos^4θ}{4}+C)
276) (displaystyle∫sin^3θdθ;u=cosθ (Hint:sin^2θ=1−cos^2θ))

(removed exercises 277- 280)

281) (displaystyle∫frac{x^2}{(x^3−3)^2}dx)

Answer:
(displaystyle−frac{1}{3(x^3−3)}+C)

In the following exercises, evaluate the definite integral.

292) (displaystyle∫^1_0xsqrt{1−x^2}dx)

293) (displaystyle∫^1_0frac{x}{sqrt{1+x^2}}dx)

Answer:
(displaystyle u=1+x^2,du=2xdx,frac{1}{2}∫^2_1u^{−1/2}du=sqrt{2}−1)

294) (displaystyle∫^2_0frac{t}{sqrt{5+t^2}}dt)

295) (displaystyle∫^1_0frac{t}{sqrt{1+t^3}}dt)

Answer:
(displaystyle u=1+t^3,du=3t^2,frac{1}{3}∫^2_1u^{−1/2}du=frac{2}{3}(sqrt{2}−1))

296) (displaystyle∫^{π/4}_0sec^2θtanθdθ)

297) (displaystyle∫^{π/4}_0frac{sinθ}{cos^4θ}dθ)

Answer:
(displaystyle u=cosθ,du=−sinθdθ,∫^1_{1/sqrt{2}}u^{−4}du=frac{1}{3}(2sqrt{2}−1))

J5.5.1)

J5.5.2)

5.6: Integrals Involving Exponential and Logarithmic Functions

In the following exercises, compute each indefinite integral.

320) (displaystyle ∫e^{2x}dx)

321) (displaystyle ∫e^{−3x}dx)

Answer:
(displaystyle frac{−1}{3}e^{−3x}+C)

322) (displaystyle ∫2^xdx)

323) (displaystyle ∫3^{−x}dx)

Answer:
(displaystyle −frac{3^{−x}}{ln3}+C)

324) (displaystyle ∫frac{1}{2x}dx)

325) (displaystyle ∫frac{2}{x}dx)

Answer:
(displaystyle ln(x^2)+C) or (displaystyle 2ln|x|+C)

326) (displaystyle ∫frac{1}{x^2}dx)

327) (displaystyle ∫frac{1}{sqrt{x}}dx)

Answer:
(displaystyle 2sqrt{x}+C)

In the following exercises, find each indefinite integral by using appropriate substitutions.

328) (displaystyle ∫frac{lnx}{x}dx)

329) (displaystyle ∫frac{dx}{x(lnx)^2})

Answer:
(displaystyle −frac{1}{lnx}+C)

336) (displaystyle ∫xe^{−x^2}dx)

337) (displaystyle ∫x^2e^{−x^3}dx)

Answer:
(displaystyle frac{−e^{−x^3}}{3}+C)

338) (displaystyle ∫e^{sinx}cosxdx)

339) (displaystyle ∫e^{tanx}sec^2xdx)

Answer:
(displaystyle e^{tanx}+C)

340) (displaystyle ∫e^{lnx}frac{dx}{x})

341) (displaystyle ∫frac{e^{ln(1−t)}}{1−t}dt)

Answer:
(displaystyle t+C)

In the following exercises, evaluate the definite integral.

355) (displaystyle ∫^2_1frac{1+2x+x^2}{3x+3x^2+x^3}dx)

Answer:
(displaystyle frac{1}{3}ln(frac{26}{7}))

356) (displaystyle ∫^{π/4}_0tanxdx)

357) (displaystyle ∫^{π/3}_0frac{sinx−cosx}{sinx+cosx}dx)

Answer:
(displaystyle ln(sqrt{3}−1))

358) (displaystyle ∫^{π/2}_{π/6}cscxdx)

359) (displaystyle ∫^{π/3}_{π/4}cotxdx)

Answer:
(displaystyle frac{1}{2}lnfrac{3}{2})

In the following exercises, integrate using the indicated substitution.

360) (displaystyle ∫frac{x}{x−100}dx;u=x−100)

361) (displaystyle ∫frac{y−1}{y+1}dy;u=y+1)

Answer:
(displaystyle y−2ln|y+1|+C)

362) (displaystyle ∫frac{1−x^2}{3x−x^3}dx;u=3x−x^3)

363) (displaystyle ∫frac{sinx+cosx}{sinx−cosx}dx;u=sinx−cosx)

Answer:
(displaystyle ln|sinx−cosx|+C)

364) (displaystyle ∫e^{2x}sqrt{1−e^{2x}}dx;u=e^{2x})

365) (displaystyle ∫ln(x)frac{sqrt{1−(lnx)^2}}{x}dx;u=lnx)

Answer:
(displaystyle −frac{1}{3}(1−(lnx^2))^{3/2}+C)

In the following exercises, (displaystyle f(x)≥0) for (displaystyle a≤x≤b). Find the area under the graph of (displaystyle f(x)) between the given values a and b by integrating.

372) (displaystyle f(x)=frac{log_{10}(x)}{x};a=10,b=100)

373) (displaystyle f(x)=frac{log_2(x)}{x};a=32,b=64)

Answer:
(displaystyle frac{11}{2}ln2)

374) (displaystyle f(x)=2^{−x};a=1,b=2)

375) (displaystyle f(x)=2^{−x};a=3,b=4)

Answer:
(displaystyle frac{1}{ln(65,536)})

376) Find the area under the graph of the function (displaystyle f(x)=xe^{−x^2}) between (displaystyle x=0) and (displaystyle x=5).

377) Compute the integral of (displaystyle f(x)=xe^{−x^2}) and find the smallest value of N such that the area under the graph (displaystyle f(x)=xe^{−x^2}) between (displaystyle x=N) and (displaystyle x=N+10) is, at most, 0.01.

Answer:
(displaystyle ∫^{N+1}_Nxe^{−x^2}dx=frac{1}{2}(e^{−N^2}−e^{−(N+1)^2}).) The quantity is less than 0.01 when (displaystyle N=2).

378) Find the limit, as N tends to infinity, of the area under the graph of (displaystyle f(x)=xe^{−x^2}) between (displaystyle x=0) and (displaystyle x=5).

379) Show that (displaystyle ∫^b_afrac{dt}{t}=∫^{1/a}_{1/b}frac{dt}{t}) when (displaystyle 0

Answer:
(displaystyle ∫^b_afrac{dx}{x}=ln(b)−ln(a)=ln(frac{1}{a})−ln(frac{1}{b})=∫^{1/a}_{1/b}frac{dx}{x})

380) Suppose that (displaystyle f(x)>0) for all x and that f and g are differentiable. Use the identity (displaystyle f^g=e^{glnf}) and the chain rule to find the derivative of (displaystyle f^g).

381) Use the previous exercise to find the antiderivative of (displaystyle h(x)=x^x(1+lnx)) and evaluate (displaystyle ∫^3_2x^x(1+lnx)dx).

Answer:
23

382) Show that if (displaystyle c>0), then the integral of (displaystyle 1/x) from ac to bc (displaystyle (0a to b.

The following exercises are intended to derive the fundamental properties of the natural log starting from the definition (displaystyle ln(x)=∫^x_1frac{dt}{t}), using properties of the definite integral and making no further assumptions.

383) Use the identity (displaystyle ln(x)=∫^x_1frac{dt}{t}) to derive the identity (displaystyle ln(frac{1}{x})=−lnx).

Solution: We may assume that (displaystyle x>1),so (displaystyle frac{1}{x}<1.) Then, (displaystyle ∫^{1/x}_{1}frac{dt}{t}). Now make the substitution (displaystyle u=frac{1}{t}), so (displaystyle du=−frac{dt}{t^2}) and (displaystyle frac{du}{u}=−frac{dt}{t}), and change endpoints: (displaystyle ∫^{1/x}_1frac{dt}{t}=−∫^x_1frac{du}{u}=−lnx.)

384) Use a change of variable in the integral (displaystyle ∫^{xy}_1frac{1}{t}dt) to show that (displaystyle lnxy=lnx+lny) for (displaystyle x,y>0).

385) Use the identity (displaystyle lnx=∫^x_1frac{dt}{x}) to show that (displaystyle ln(x)) is an increasing function of x on (displaystyle [0,∞)), and use the previous exercises to show that the range of (displaystyle ln(x)) is (displaystyle (−∞,∞)). Without any further assumptions, conclude that (displaystyle ln(x)) has an inverse function defined on (displaystyle (−∞,∞).)

386) Pretend, for the moment, that we do not know that (displaystyle e^x) is the inverse function of (displaystyle ln(x)), but keep in mind that (displaystyle ln(x)) has an inverse function defined on (displaystyle (−∞,∞)). Call it E. Use the identity (displaystyle lnxy=lnx+lny) to deduce that (displaystyle E(a+b)=E(a)E(b)) for any real numbers a, b.

387) Pretend, for the moment, that we do not know that (displaystyle e^x) is the inverse function of (displaystyle lnx), but keep in mind that (displaystyle lnx) has an inverse function defined on (displaystyle (−∞,∞)). Show that (displaystyle E'(t)=E(t).)

Solution: (displaystyle x=E(ln(x)).) Then, (displaystyle 1=frac{E'(lnx)}{x}) or (displaystyle x=E'(lnx)). Since any number t can be written (displaystyle t=lnx) for some x, and for such t we have (displaystyle x=E(t)), it follows that for any (displaystyle t,E'(t)=E(t).)

388) The sine integral, defined as (displaystyle S(x)=∫^x_0frac{sint}{t}dt) is an important quantity in engineering. Although it does not have a simple closed formula, it is possible to estimate its behavior for large x. Show that for (displaystyle k≥1,|S(2πk)−S(2π(k+1))|≤frac{1}{k(2k+1)π}.) (Hint: (displaystyle sin(t+π)=−sint))

389) [T] The normal distribution in probability is given by (displaystyle p(x)=frac{1}{σsqrt{2π}}e^{−(x−μ)^2/2σ^2}), where σ is the standard deviation and μ is the average. The standard normal distribution in probability, (displaystyle p_s), corresponds to (displaystyle μ=0) and (displaystyle σ=1). Compute the left endpoint estimates (displaystyle R_{10}) and (displaystyle R_{100}) of (displaystyle ∫^1_{−1}frac{1}{sqrt{2π}}e^{−x^{2/2}}dx.)

Solution: (displaystyle R_{10}=0.6811,R_{100}=0.6827)

390) [T] Compute the right endpoint estimates (displaystyle R_{50}) and (displaystyle R_{100}) of (displaystyle ∫^5_{−3}frac{1}{2sqrt{2π}}e^{−(x−1)^2/8}).


Groundwater flow system analysis in the regolith of Dodowa on the Accra Plains, Ghana

The Togo Structural Unit is weathered into a saprolite of at most 50 m thickness composed of both unconfined and confined parts of a shallow aquifer.

In the Dahomeyan Structural Unit, groundwater was encountered in fractures under both confined and unconfined conditions.

Infiltration of waste water gave rise to polluted groundwater with elevated electrical conductivity values and high nitrate concentrations.

Low transmissivities (<5.8e–5 m2/s (=5 m2/d)) restrict groundwater use to small scale supply.

Groundwater system development in the area is confined to the Dodowa area.


Q: Explain, using limits, why f(x) = - is not continuous at x = 0.

A: To Determine: explain using limits why f(x)=1x is not continuous at x=0. Given: we have a functionf.

A: Click to see the answer

Q: (a) Estimate the doubling time of the exponential function shown in the figure below. P 240 180 120 .

A: Given that the graph of exponential function: From the graph of exponential function, at t=0, P=.

Q: Given f(x) = 3x3 + 4x2 - 5x + 3, use the Remainder Theorem to find f(-4).

A: Click to see the answer

A: The function f(x) is, f(x)=x2+5xx3-25x=x(x+5)x(x2-52)=x(x+5)x(x-5)(x+5)=1x-5 The denominator of the .

Q: Find the derivative of the function, arcsin 3r a. # 45. g(x)

A: since you have asked multiple questions in a single request, we would be answering only first questi.

Q: If f(8)=4, f'(8)--6, g(8)--4, and g'(8)-1 and h(x)-5f(x)-3g(x)

A: Click to see the answer

Q: Give the starting value a, the growth rate r, and the continuous growth rate k. Q = 12.3 · 10-0.15 t.

A: Given: The equation Q=12.3⋅10−0.15t .

Q: Convert Q = 5e" to the form Q = ab' . Round your answer for b to three decimal places. Q =

A: We will use the formula: xmn=xmn in reverse. Q=5e6tQ=5(e6)t Q=5(403.429)t Comparing with Q=abt, we .


Q: Explain, using limits, why f(x) = - is not continuous at x = 0.

A: To Determine: explain using limits why f(x)=1x is not continuous at x=0. Given: we have a functionf.

A: Click to see the answer

Q: (a) Estimate the doubling time of the exponential function shown in the figure below. P 240 180 120 .

A: Given that the graph of exponential function: From the graph of exponential function, at t=0, P=.

Q: Given f(x) = 3x3 + 4x2 - 5x + 3, use the Remainder Theorem to find f(-4).

A: Click to see the answer

A: The function f(x) is, f(x)=x2+5xx3-25x=x(x+5)x(x2-52)=x(x+5)x(x-5)(x+5)=1x-5 The denominator of the .

Q: Find the derivative of the function, arcsin 3r a. # 45. g(x)

A: since you have asked multiple questions in a single request, we would be answering only first questi.

Q: If f(8)=4, f'(8)--6, g(8)--4, and g'(8)-1 and h(x)-5f(x)-3g(x)

A: Click to see the answer

Q: Give the starting value a, the growth rate r, and the continuous growth rate k. Q = 12.3 · 10-0.15 t.

A: Given: The equation Q=12.3⋅10−0.15t .

Q: Convert Q = 5e" to the form Q = ab' . Round your answer for b to three decimal places. Q =

A: We will use the formula: xmn=xmn in reverse. Q=5e6tQ=5(e6)t Q=5(403.429)t Comparing with Q=abt, we .


5.2 Frequency

As mentioned above, the term pitch refers to the “highness” or “lowness” of a particular tone. The shrill whistling of a tea kettle is an example of a high pitch. The deep, resonating horn of an enormous freighter ship is an example of a low pitch. It is important to remember, though, that highness and lowness are relative—this is particularly important when describing musical tones, where subtle changes in pitch can have dramatic effects on a listener’s experience. A high-pitched musical tone, in other words, may be described as low when compared to another, even higher pitched tone.

The terms “highness” and “lowness” are quite common in discussions of pitch. The image they suggest—of pitches placed along a vertical axis in physical space—is, however, just an analogy. When we talk about the high speed of a train, we are not referring to the elevation of the tracks and in music there is nothing inherent to a high pitch that places it physically above any other. Nonetheless, the vertical imagery is helpful, particularly when it comes to the way pitches are written in staff notation, as we will see momentarily.

Although a detailed discussion of musical acoustics is beyond the scope of this book, we may define pitch more accurately by considering the physical phenomena that produce sound. When an object vibrates, it sets the air around it into motion. The air molecules are compressed and decompressed in correspondence with the motion of the vibrating object. These tiny waves of pressure then emanate outward, away from their source. Human ears are capable of perceiving these vibrations in the air as sounds. If the pulses of compression happen regularly, they will be perceived as having pitch.

Pitch corresponds with the frequency of these vibrations: objects producing high pitches vibrate very quickly, objects producing low pitches less so. Pitch is measured in hertz (Hz), a unit indicating the number of vibrations happening over a time span of one second. Example 5–1 presents a 440 Hz tone, a pitch produced by vibrations happening 440 times every second:

Changing the frequency of the vibrations changes the pitch. When the vibrations happen more frequently, we perceive a higher pitch. Example 5–2 presents a 493.88 Hz tone. It sounds slightly “higher”—more urgent or energetic—than the tone in Example 5–1.

Listen to each of the following pairs of pitches and determine which of the two is higher. (You will need to view this chapter in the online version of this book to complete this activity.)

Exercise 5–1a:

Question

Which of the following pitches is higher, the first or second?

Does the pitch seem to rise or fall from the first note to the second?

The first pitch is higher.

Exercise 5–1b:

Question

Which of the following pitches is higher, the first or second?

Does the pitch seem to rise or fall from the first note to the second?

The second pitch is higher.

Exercise 5–1c:

Question

Which of the following pitches is higher, the first or second?

Does the pitch seem to rise or fall from the first note to the second?

The second pitch is higher.

Exercise 5–1d:

Question

Which of the following pitches is higher, the first or second?

Does the pitch seem to rise or fall from the first note to the second?

The first pitch is higher.

Exercise 5–1e:

Question

Which of the following pitches is higher, the first or second?

Does the pitch seem to rise or fall from the first note to the second?

The first pitch is higher.

Exercise 5–1f:

Question

Which of the following pitches is higher, the first or second?

Does the pitch seem to rise or fall from the first note to the second?

The second pitch is higher.

There are an infinite number of pitches. It follows, then, that there are also an infinite number of pitches between any two pitches. Some pitches are so close in frequency that it is impossible to discern the difference between them. Furthermore, some pitches are either so high or so low in frequency that they are imperceptible to the human ear. Generally speaking, humans are capable of hearing pitches in the 20 Hz to 20,000 Hz range. In tonal Western art music, though, the pitches one encounters tend to be much more limited in range and in number.


Leveling Up

As your character goes on adventures and overcomes challenges, he or she gains experience, represented by experience points. A character who reaches a specified experience point total advances in capability. This advancement is called gaining a level.

When your character gains a level, his or her class often grants additional features, as detailed in the class description. Some of these features allow you to increase your ability scores, either increasing two scores by 1 each or increasing one score by 2. You can’t increase an ability score above 20. In addition, every character's proficiency bonus increases at certain levels.

Each time you gain a level, you gain 1 additional Hit Die. Roll that Hit Die, add your Constitution modifier to the roll, and add the total to your hit point maximum. Alternatively, you can use the fixed value shown in your class entry, which is the average result of the die roll (rounded up).

When your Constitution modifier increases by 1, your hit point maximum increases by 1 for each level you have attained. For example, if your 7th-level fighter has a Constitution score of 18, when he reaches 8th level, he increases his Constitution score from 17 to 18, thus increasing his Constitution modifier from +3 to +4. His hit point maximum then increases by 8.

The Character Advancement table summarizes the XP you need to advance in levels from level 1 through level 20, and the proficiency bonus for a character of that level. Consult the information in your character's class description to see what other improvements you gain at each level.


5.5E & 5.6E U-Substitution Exercises

SOLUTION 10 : Integrate . Use u-substitution. Let

so that (Don't forget to use the chain rule on e - x .)

du = 3 e - x (-1) dx = -3 e - x dx ,

However, how can we replace the term e -3 x in the original problem ? Note that

we can "back substitute" with

Substitute into the original problem, replacing all forms of x , getting

(Recall that ( AB ) C = A C B C .)

Click HERE to return to the list of problems.

SOLUTION 11 : Integrate . Use u-substitution. Let

so that (Don't forget to use the chain rule on e 2 x .)

Substitute into the original problem, replacing all forms of x , getting

(Do not make the following VERY COMMON MISTAKE : . Why is this INCORRECT ?)

Click HERE to return to the list of problems.

SOLUTION 12 : Integrate . First, factor out e 9 x from inside the parantheses. Then

(Recall that ( AB ) C = A C B C .)

(Recall that ( A B ) C = A BC .)

Now use u-substitution. Let

so that (Don't forget to use the chain rule on e 3 x .)

Substitute into the original problem, replacing all forms of x , and getting


5.5E & 5.6E U-Substitution Exercises

We now need to go back and revisit the substitution rule as it applies to definite integrals. At some level there really isn’t a lot to do in this section. Recall that the first step in doing a definite integral is to compute the indefinite integral and that hasn’t changed. We will still compute the indefinite integral first. This means that we already know how to do these. We use the substitution rule to find the indefinite integral and then do the evaluation.

There are however, two ways to deal with the evaluation step. One of the ways of doing the evaluation is the probably the most obvious at this point, but also has a point in the process where we can get in trouble if we aren’t paying attention.

Let’s work an example illustrating both ways of doing the evaluation step.

Let’s start off looking at the first way of dealing with the evaluation step. We’ll need to be careful with this method as there is a point in the process where if we aren’t paying attention we’ll get the wrong answer.

We’ll first need to compute the indefinite integral using the substitution rule. Note however, that we will constantly remind ourselves that this is a definite integral by putting the limits on the integral at each step. Without the limits it’s easy to forget that we had a definite integral when we’ve gotten the indefinite integral computed.

In this case the substitution is,

Plugging this into the integral gives,

Notice that we didn’t do the evaluation yet. This is where the potential problem arises with this solution method. The limits given here are from the original integral and hence are values of (t). We have (u)’s in our solution. We can’t plug values of (t) in for (u).

Therefore, we will have to go back to (t)’s before we do the substitution. This is the standard step in the substitution process, but it is often forgotten when doing definite integrals. Note as well that in this case, if we don’t go back to (t)’s we will have a small problem in that one of the evaluations will end up giving us a complex number.

So, finishing this problem gives,

So, that was the first solution method. Let’s take a look at the second method.

Note that this solution method isn’t really all that different from the first method. In this method we are going to remember that when doing a substitution we want to eliminate all the (t)’s in the integral and write everything in terms of (u).

When we say all here we really mean all. In other words, remember that the limits on the integral are also values of (t) and we’re going to convert the limits into (u) values. Converting the limits is pretty simple since our substitution will tell us how to relate (t) and (u) so all we need to do is plug in the original (t) limits into the substitution and we’ll get the new (u) limits.

Here is the substitution (it’s the same as the first method) as well as the limit conversions.

[eginu & = 1 - 4hspace <0.25in>du = - 12dthspace <0.25in>Rightarrow hspace <0.25in>dt = - frac<1><<12>>du t & = - 2hspace <0.5in>Rightarrow hspace<0.5in>u = 1 - 4 ight)^3> = 33 t & = 0hspace <0.65in>Rightarrow hspace<0.5in>u = 1 - 4 = 1end]

As with the first method let’s pause here a moment to remind us what we’re doing. In this case, we’ve converted the limits to (u)’s and we’ve also got our integral in terms of (u)’s and so here we can just plug the limits directly into our integral. Note that in this case we won’t plug our substitution back in. Doing this here would cause problems as we would have (t)’s in the integral and our limits would be (u)’s. Here’s the rest of this problem.

We got exactly the same answer and this time didn’t have to worry about going back to (t)’s in our answer.

So, we’ve seen two solution techniques for computing definite integrals that require the substitution rule. Both are valid solution methods and each have their uses. We will be using the second almost exclusively however since it makes the evaluation step a little easier.

Let’s work some more examples.

  1. ( displaystyle int_<<, - 1>>^<<,5>>< ight)<> ight)>^5>,dw>>)
  2. ( displaystyle int_<<, - 2>>^<<, - 6>><<<< ight)>^3>>> - frac<5><<1 + 2x>>,dx>>)
  3. ( displaystyle int_<<,0>>^<<,frac<1><2>>><<<<f>^y> + 2cos left( ight),dy>>)
  4. ( displaystyle int_<<,frac<3>>>^<<,0>><<3sin left( <2>> ight) - 5cos left( ight),dz>>)

Since we’ve done quite a few substitution rule integrals to this time we aren’t going to put a lot of effort into explaining the substitution part of things here.

The substitution and converted limits are,

Sometimes a limit will remain the same after the substitution. Don’t get excited when it happens and don’t expect it to happen all the time.

Don’t get excited about large numbers for answers here. Sometimes they are. That’s life.

Here is the substitution and converted limits for this problem,

This integral needs to be split into two integrals since the first term doesn’t require a substitution and the second does.

Here is the substitution and converted limits for the second term.

This integral will require two substitutions. So first split up the integral so we can do a substitution on each term.

There are the two substitutions for these integrals.

Here is the integral for this problem.

The next set of examples is designed to make sure that we don’t forget about a very important point about definite integrals.

Be careful with this integral. The denominator is zero at (t = pm frac<1><2>) and both of these are in the interval of integration. Therefore, this integrand is not continuous in the interval and so the integral can’t be done.

Be careful with definite integrals and be on the lookout for division by zero problems. In the previous section they were easy to spot since all the division by zero problems that we had there were where the variable was itself zero. Once we move into substitution problems however they will not always be so easy to spot so make sure that you first take a quick look at the integrand and see if there are any continuity problems with the integrand and if they occur in the interval of integration.

Now, in this case the integral can be done because the two points of discontinuity, (t = pm frac<1><2>), are both outside of the interval of integration. The substitution and converted limits in this case are,

Let’s work another set of examples. These are a little tougher (at least in appearance) than the previous sets.

  1. ( displaystyle int_<<,0>>^ <<,ln left( <1 + pi > ight)>><<<<f>^x>cos left( <1 - <<f>^x>> ight)>>,dx)
  2. ( displaystyle int_<<<<f>^2>>>^<<<<f>^6>>>< ight]>^4>>>,dt>>)
  3. ( displaystyle int_<<<12>>>>^<<,frac<9>>>< ight) an left( <3P> ight)>>< ight)>>>>,dP>>)
  4. ( displaystyle int_<<, - pi >>^<<,frac<2>>>< ight),dx>>)
  5. ( displaystyle int_<<<50>>>>^<2><>^>>>><<>>,dw>>)

The limits are a little unusual in this case, but that will happen sometimes so don’t get too excited about it. Here is the substitution.

Here is the substitution and converted limits for this problem.

Here is the substitution and converted limits and don’t get too excited about the substitution. It’s a little messy in the case, but that can happen on occasion.

[eginu & = 2 + sec left( <3P> ight),,,,,,,,du = 3sec left( <3P> ight) an left( <3P> ight)dP,,,,, Rightarrow ,,,,,sec left( <3P> ight) an left( <3P> ight)dP = frac<1><3>du P & = frac<<12>>hspace <0.5in>Rightarrow hspace<0.25in>u = 2 + sec left( <4>> ight) = 2 + sqrt 2 P & = frac<9>hspace <0.5in>Rightarrow hspace<0.25in>u = 2 + sec left( <3>> ight) = 4end]

So, not only was the substitution messy, but we also have a messy answer, but again that’s life on occasion.

This problem not as bad as it looks. Here is the substitution and converted limits.

[eginu & = sin xhspace<0.25in>du = cos x,dx x & = frac<2>hspace<0.25in>,,,, Rightarrow hspace<0.25in>,u = sin frac <2>= 1hspace<0.5in>x = - pi hspace<0.25in>,,,,,,, Rightarrow hspace<0.25in>,,,,,,u = sin left( < - pi > ight) = 0end]

The cosine in the very front of the integrand will get substituted away in the differential and so this integrand actually simplifies down significantly. Here is the integral.

Don’t get excited about these kinds of answers. On occasion we will end up with trig function evaluations like this.

This is also a tricky substitution (at least until you see it). Here it is,

In this last set of examples we saw some tricky substitutions and messy limits, but these are a fact of life with some substitution problems and so we need to be prepared for dealing with them when they happen.


V1.5 FPDS-NG Data Dictionary

A requirements contract provides for filling all actual purchase requirements of designated Government activities for supplies or services during a specified contract period, with deliveries or performance to be scheduled by placing orders with the contractor. A Requirements IDC or Multi-Agency Contract is a contract for all of the agency's requirement for the supplies or services specified, and effective for the period stated, in the IDC or Multi-Agency Contract. After award, the contract is a mandatory source for the agency for the supplies or services specified. The quantities of supplies or services specified in the IDC or Multi-Agency Contract are estimates only and are not purchased by this contract. Except as this contract may otherwise provide, if the Government's requirements do not result in orders in the quantities described as "estimated" or "maximum" in the Schedule, that fact shall not constitute the basis for an equitable price adjustment.

An indefinite-quantity contract provides for an indefinite quantity, within stated limits, of supplies or services during a fixed period. The Government places orders for individual requirements. Quantity limits may be stated as number of units or as dollar values. An Indefinite-Quantity is a contract for the supplies or services specified, and effective for the period stated, in the IDC or Multi-Agency Contract. The quantities of supplies and services specified in the IDC or Multi-Agency Contract are estimates only and are not purchased by this contract.

A definite-quantity contract provides for delivery of a definite quantity of specific supplies or services for a fixed period, with deliveries or performance to be scheduled at designated locations upon order. A Definite Quantity IDC or Multi-Agency Contract is a definite-quantity, indefinite-delivery contract for the supplies or services specified, and effective for the period stated, in the IDC or Multi-Agency Contract.


5.5E & 5.6E U-Substitution Exercises

COM HO ESCRIVIM JOGUINES O JUGUINES?

COM HO ESCRIVI M TURRÓ O TORRÓ?

La coincidència de so de la "o" i de la "u" àtones (NO FORTES ) origina dubtes, que es poden resoldre si se seguieixen les regles següents:

Si el so de la [u] a l'ùltima síl·laba de la paraula. TOR [U] que escrivim U o O.

Els substantius (noms) masculins que acabin en [u] normalment s-escriuen amb -o:

  • exemples: gerro, toro suro carro
  • excepcions: museu correu tribu

Els noms i adjectius que fan el plural en SO [us] normalment s-escriuen amb -os final:

exemples: excepcions:
abús -> abusos museu -> museus
gras -> grassos actiu -> actius
bosc -> boscos motiu -> motius
feliç -> feliços europeu -> europeus
anis -> anissos

POSEM U en els noms invariables acabats en -us:

POSEM U en les paraules acabades en diftong decreixent (au, eu, iu, ou):

POSEM O a la primera persona del singular del present d’indicatiu:

A l'interior de la paraula.

Per saber si hem d’escriur e "O " o " U" , quan es troben al mig d’una paraula, cal que busquem una paraula de la mateixa família que tingui la lletra (vo c al) en posició tònica.(forta)

S'escriu O
Perquè prove del derivat de: S'escriu U
Perquè prové del derivat de:
foscor fosc duresa dur
novè nou llunyà lluny
pomera poma gruixut gruix
boirós boire fuster fusta

català amb O / cast. amb U

atordir, atorrollar, atribolar (atribolat -ada) o tribular (tribulació), avorrir (avorriment), bordell, botifarra,
brúixola, calorós -osa, capítol, cartolina, colobra, complir, cònsol (consolat), corbat -ada, embotir, escapolirse, escrúpol, esdrúixol -a, gola (engolir), Hongria (hongarès -esa), Joan, joglar joglaressa, joguet, joguina, Josep, joventut, muntar (i der., muntatge), nodrir, ombria, ordir, pèndol, ploma (plomatge, plomar), podrir, polir, pols (polsera, polsació), rigorós -osa, robí, roí roïna, Romania, rossinyol, sofrir, solc (solcar), sorgir, sorgir (ressorgir), sospir (sospirar), sostraure o sostreure, tamboret, tenidoria, títol, tomba (ultratomba), tonyina, torba (torbar, torbació, torbador -ora), torró, torticoli, triomf .

català amb U / cast. amb O


ateneu, bufetada, butlletí, cacau, continu -ínua, correu, cuirassa, escullera, fetus, focus, fòrum, individu,
ingenu -ènua, muntanya, porus, ritu, saurí saurina, sèrum, suborn (subornar), sufocar, supèrbia, suport,
(suportar, insuportable), tètanus, tramuntana, trofeu, turment (turmentar)


Watch the video: MSML2020 Paper Presentation - Xueshuang Xiang (May 2022).