I am having difficulty understanding how to use the differential of du dx. In problem 2 above, how did 15x^2 turn to be 5*3*x^2?

They are equivalent is the short answer. Long answer is, since u was chosen to be x^3 - 7 the derivative is 3x^2, so we want something to substitute that in for. really the important part is x^2. as long as that exists you can "factor out" a 1/3 and then you'll have 1/3 * 3x^2 and have what you need to substitute. Luckily, there is 15x^2, and if you factor out a 5 you get the 3x^2 you need, which is where 5*3x^2 came from. Does that help? if not I can maybe try walking through it differently than the problem.

Could someone please explain how ∫sec(u)tan(u)du =sec(u)+C Tnx!

The identity sec x d/dx = sec(x)*tan(x) is well known. So the answer is just reverse chain rule.

Main content

𝘶-substitution with definite integrals

Google Classroom

Performing u-substitution with definite integrals is very similar to how it's done with indefinite integrals, but with an added step: accounting for the limits of integration. Let's see what this means by finding integral, start subscript, 1, end subscript, squared, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, x, squared, plus, 1, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, x, end color #7854ab.

We notice that start color #7854ab, 2, x, end color #7854ab is the derivative of start color #1fab54, x, squared, plus, 1, end color #1fab54, so u-substitution applies. Let start color #1fab54, u, equals, x, squared, plus, 1, end color #1fab54, then start color #7854ab, d, u, equals, 2, x, d, x, end color #7854ab. Now we substitute:

integral, start subscript, 1, end subscript, squared, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, x, squared, plus, 1, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, x, end color #7854ab, equals, integral, start subscript, 1, end subscript, squared, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, u, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, u, end color #7854ab

Wait a minute! The limits of integration were fitted for x, not for u. Think about this graphically. We wanted the area under the curve start color #11accd, y, equals, 2, x, left parenthesis, x, squared, plus, 1, right parenthesis, cubed, end color #11accd between x, equals, 1 and x, equals, 2.

Now that we changed the curve to start color #aa87ff, y, equals, u, cubed, end color #aa87ff, why should the limits stay the same?

Both start color #11accd, y, equals, 2, x, left parenthesis, x, squared, plus, 1, right parenthesis, cubed, end color #11accd and start color #aa87ff, y, equals, u, cubed, end color #aa87ff are graphed. You can see the areas under the curves between x, equals, 1 and x, equals, 2 (or u, equals, 1 and u, equals, 2) are very different in size.

Indeed, the limits shouldn't stay the same. To find the new limits, we need to find what values of start color #1fab54, u, end color #1fab54 correspond to start color #1fab54, x, squared, plus, 1, end color #1fab54 for x, equals, start color #ca337c, 1, end color #ca337c and x, equals, start color #ca337c, 2, end color #ca337c:

Lower bound: left parenthesis, start color #ca337c, 1, end color #ca337c, right parenthesis, squared, plus, 1, equals, start color #ca337c, 2, end color #ca337c
Upper bound: left parenthesis, start color #ca337c, 2, end color #ca337c, right parenthesis, squared, plus, 1, equals, start color #ca337c, 5, end color #ca337c

Now we can correctly perform the u-substitution:

integral, start subscript, start color #ca337c, 1, end color #ca337c, end subscript, start superscript, start color #ca337c, 2, end color #ca337c, end superscript, start color #7854ab, 2, x, end color #7854ab, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, x, squared, plus, 1, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, x, end color #7854ab, equals, integral, start subscript, start color #ca337c, 2, end color #ca337c, end subscript, start superscript, start color #ca337c, 5, end color #ca337c, end superscript, start color #e07d10, left parenthesis, end color #e07d10, start color #1fab54, u, end color #1fab54, start color #e07d10, right parenthesis, cubed, end color #e07d10, start color #7854ab, d, u, end color #7854ab

start color #aa87ff, y, equals, u, cubed, end color #aa87ff is graphed with the area from u, equals, 2 to u, equals, 5. Now we can see that the shaded areas look roughly the same size (they are actually exactly the same size, but it's hard to say by just looking).

From here on, we can solve everything according to u:

\begin{aligned} \displaystyle\int_{2}^5 u^3\,du&=\left[\dfrac{u^4}{4}\right]_{2}^5 \\\\ &=\dfrac{5^4}{4}-\dfrac{2^4}{4} \\\\ &=152.25 \end{aligned}

[Is this the only way to account for the limits of integration?]

Remember: When using u-substitution with definite integrals, we must always account for the limits of integration.

Problem 1

Ella was asked to find integral, start subscript, 1, end subscript, start superscript, 5, end superscript, left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, squared, plus, x, right parenthesis, cubed, d, x. This is her work:

Step 1: Let u, equals, x, squared, plus, x

Step 2: d, u, equals, left parenthesis, 2, x, plus, 1, right parenthesis, d, x

Step 3:

integral, start subscript, 1, end subscript, start superscript, 5, end superscript, left parenthesis, 2, x, plus, 1, right parenthesis, left parenthesis, x, squared, plus, x, right parenthesis, cubed, d, x, equals, integral, start subscript, 1, end subscript, start superscript, 5, end superscript, u, cubed, d, u

Step 4:

\begin{aligned} \displaystyle\int_1^5 u^3du&=\left[\dfrac{u^4}{4}\right]_1^5 \\\\ &=\dfrac{5^4}{4}-\dfrac{1^4}{4} \\\\ &=156 \end{aligned}

Is Ella's work correct? If not, what is her mistake?

(Choice A)
Ella's work is correct.
(Choice B)
Step 1 is incorrect. Ella should've defined u as 2, x, plus, 1.
(Choice C)
Step 3 is incorrect. Ella didn't change the limits of integration.
(Choice D)
Step 4 is incorrect. The antiderivative of u, cubed is 3, u, squared.

Problem 2

integral, start subscript, 1, end subscript, squared, 15, x, squared, left parenthesis, x, cubed, minus, 7, right parenthesis, start superscript, 4, end superscript, d, x, equals, question mark

Want more practice? Try this exercise.

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vanessa hili
Posted 3 years ago. Direct link to vanessa hili's post “I am having difficulty un...”
I am having difficulty understanding how to use the differential of du dx. In problem 2 above, how did 15x^2 turn to be 5*3*x^2?
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(1 vote)
Answer
- loumast17
  Posted 3 years ago. Direct link to loumast17's post “They are equivalent is th...”
  They are equivalent is the short answer.
  
  Long answer is, since u was chosen to be x^3 - 7 the derivative is 3x^2, so we want something to substitute that in for. really the important part is x^2. as long as that exists you can "factor out" a 1/3 and then you'll have 1/3 * 3x^2 and have what you need to substitute.
  
  Luckily, there is 15x^2, and if you factor out a 5 you get the 3x^2 you need, which is where 5*3x^2 came from.
  
  Does that help? if not I can maybe try walking through it differently than the problem.
  Button navigates to signup page
  (5 votes)
Kaylagrey
Posted a year ago. Direct link to Kaylagrey's post “Could someone please expl...”
Could someone please explain how

∫sec(u)tan(u)du
=sec(u)+C

Tnx!
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(1 vote)
Answer
- cossine
  Posted a year ago. Direct link to cossine's post “The identity sec x d/dx =...”
  The identity sec x d/dx = sec(x)*tan(x) is well known. So the answer is just reverse chain rule.
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  (2 votes)
orel7403
Posted 2 years ago. Direct link to orel7403's post “what if the bounds get fl...”
what if the bounds get flipped after changing them as in you would be integrating from a to b when b<a can you still leave the u sub or do you have to swap them
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(1 vote)
Answer
- Jeff
  Posted 2 years ago. Direct link to Jeff's post “If the bounds become inve...”
  If the bounds become inverted (b<a) due to a u-sub, it is typically best switch them back. It is OK to switch the bounds as long as you add a negative out front of the integral to make up for it.
  
  If you don't fix the "backwards bounds" you will still end up with the same answer in the end. However, it's good practice to swap them and add the negative.
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  (1 vote)
石乐志大师
Posted 10 months ago. Direct link to 石乐志大师's post “Why does the definite int...”
Why does the definite integral of u^3 from 2 to 5 has the same area as the definite integral of 2x(x^2+1)^3 from 1 to 2?
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(1 vote)
Answer
- cossine
  Posted 10 months ago. Direct link to cossine's post “So I am assuming you have...”
  So I am assuming you have used u-substitution. You can look up the the theorem to confirm it works.
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  (1 vote)
gunank312
Posted 3 years ago. Direct link to gunank312's post “i tried to plot graph of ...”
i tried to plot graph of 2x(x^2 + 1)^3 and (x^2 + 1)^3 , but why am i not getting the same graphs as depicted? we are trying to find area under curve given the bounds only right?
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(1 vote)
Answer
- Road to 1 Million Energy Points!
  Posted 3 years ago. Direct link to Road to 1 Million Energy Points!'s post “What you have there is th...”
  What you have there is the anti-derivative, not the definite integral, so it is a completely separate graph entirely
  Comment on Road to 1 Million Energy Points!'s post “What you have there is th...”
  (1 vote)
escapingperil
Posted 3 years ago. Direct link to escapingperil's post “I am working on a questio...”
I am working on a question like this and having trouble. It is asking me to find the area between the curve y=xsqrt(4-x^2) and the x-axis over the interval [-2, 2].

I am mostly confused on how not to get zero as an answer. Should I break it into multiple parts?
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(1 vote)
Answer
- Jerry Nilsson
  Posted 3 years ago. Direct link to Jerry Nilsson's post “𝑓(𝑥) = 𝑥√(4 − 𝑥²) 𝑥...”
  𝑓(𝑥) = 𝑥√(4 − 𝑥²)
  
  𝑥 ∈ (−2, 0) ⇔ 𝑓(𝑥) < 0
  Thus, the (positive) area is
  𝐴 = ∫[0, 2] 𝑓(𝑥)𝑑𝑥 − ∫[−2, 0] 𝑓(𝑥)𝑑𝑥
  
  However, because 𝑓(𝑥) is an odd function, we can just write
  𝐴 = 2 ∙ ∫[0, 2] 𝑓(𝑥)𝑑𝑥
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  (1 vote)
caiyresl
Posted 3 years ago. Direct link to caiyresl's post “Can you substitute the or...”
Can you substitute the original expression with x for u and then use the same boundaries? In the example, would (x^2+1)^4/4 for [1,2] work?
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(1 vote)
Answer
- Misha Taylor
  Posted 3 years ago. Direct link to Misha Taylor's post “Exactly. It's easier that...”
  Exactly. It's easier that way. The Limits might get too big. With Polynomials and cubic and quintic equations more shortcuts are ready, since (a+b)^5=a^5+b^5+5ab(a+b)^3+5ab(a+b), and One can use it more if the have a quintic nested cubic or vice versa, when doing crazy stuff with diffEQ and regression to a smaller diffEQ that's easier. If You're dealing with a+b, rather have a^5+b^5 or (a+b)^5 be ready. If a^5+b^5=x, and a+b=y, and ab=c. I'd think of something like x=y^5-5cy^3+5cy; You put in the limits, it isn't as ready for running parallel processes. Running parallel processes in Math is the Best Part of ADHD and Autism, and it finds out more information when connecting the dots gets connected. If You don't run enough parallel processes, you might not find the quickest solution. For Some Reason, 42*6, can be rewritten as 36*7, faster than multiplying 6*7, because parallel processes were "strong as a carbon atom". Multiplying 512*84, can be quickly rewritten as 42000+24*42, which can be quickly done be rewritten it again as 12*84 which->1200-12*16->1200-(196-16), and so on. Verify that by another parallel process like 4096*12 which is 480000+1200-48 which is probably 481152. If That's wrong, just create another parallel process. Verify it by thinking does 4*2 end in 4. Nope! So 2048*21=42000+1000-((1*48-2*21=>6)+2=>8)->43008. Or 1050-42->8+43000->43008. Run More Threads and so on.
  Comment on Misha Taylor's post “Exactly. It's easier that...”
  (1 vote)

AP®︎ Calculus BC (2017 edition)

Course: AP®︎ Calculus BC (2017 edition) > Unit 9

𝘶-substitution with definite integrals

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