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.

Alright, I really wanted to see the difference in bounds graphically, and this is nice. However, I am confused. If (u^3)(du) = 2x(x^2+1) dx, then how are the graphs different? I thought du and dx were both just the bases of an infinite number of rectangles and they both approached zero. What actually happens when we substitute a variable or when we take the integral with respect to a different variable? I thought they would be equivalent, but now I'm more confused.

Good question! I'll try to put in a simple manner. So, we have two variables here: x and u. Think of it as two worlds: the (x,f(x)) worlds (where input is x and output is f(x)) and the (u,f(u)) world (where input is u and output is f(u)). Now, the graph will look different in both these worlds. What a u-substitution does is that it creates a map from the x world to the u world (i.e. the substitution we make maps every value of x to a corresponding value of u). As a result, every point is mapped onto a new coordinate system where u = x^2 + 1. This makes our graph into something we can easily find the area under. Also, a more solid reason as to why the graphs will be different is because of the substitution. See that u = x^2 + 1. If u = x, then the graphs would be the same. But, u = x is a useless substitution, as it just changes the variable and doesn't make anything easier for us. If you can understand this idea, and you choose to pursue calculus in the future, this visualization helps tremendously when you reach multivariable calculus.

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.

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?

Yes, after taking the antiderivative with respect to 𝑢, we can substitute back to get the antiderivative as a function of 𝑥 and evaluate that for the original boundaries.

Main content

Integral Calculus

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

\int_{1}^{2} 2 x (x^{2} + 1)^{3} d x

We notice that

2 x

is the derivative of

x^{2} + 1

, so

u

-substitution applies. Let

u = x^{2} + 1

, then

d u = 2 x d x

. Now we substitute:

\int_{1}^{2} 2 x (x^{2} + 1)^{3} d x = \int_{1}^{2} (u)^{3} d u

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

y = 2 x (x^{2} + 1)^{3}

between

x = 1

and

x = 2

Now that we changed the curve to

y = u^{3}

, why should the limits stay the same?

Indeed, the limits shouldn't stay the same. To find the new limits, we need to find what values of

u

correspond to

x^{2} + 1

for

x = 1

and

x = 2

Lower bound: $(1)^{2} + 1 = 2$ ‍
Upper bound: $(2)^{2} + 1 = 5$ ‍

Now we can correctly perform the

u

-substitution:

\int_{1}^{2} 2 x (x^{2} + 1)^{3} d x = \int_{2}^{5} (u)^{3} d u

y = u^{3}

is graphed with the area from

u = 2

u = 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} \int_{2}^{5} u^{3} d u & = {[\frac{u^{4}}{4}]}_{2}^{5} \\ = \frac{5^{4}}{4} - \frac{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

\int_{1}^{5} (2 x + 1) (x^{2} + x)^{3} d x

. This is her work:

Step 1: Let

u = x^{2} + x

Step 2:

d u = (2 x + 1) d x

Step 3:

\int_{1}^{5} (2 x + 1) (x^{2} + x)^{3} d x = \int_{1}^{5} u^{3} d u

Step 4:

\begin{aligned} \int_{1}^{5} u^{3} d u & = {[\frac{u^{4}}{4}]}_{1}^{5} \\ = \frac{5^{4}}{4} - \frac{1^{4}}{4} \\ = 156 \end{aligned}

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

Problem 2

\int_{1}^{2} 15 x^{2} (x^{3} - 7)^{4} d x = ?

Want more practice? Try this exercise.

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vanessa hili
Posted 4 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?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- loumast17
  Posted 4 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
  (7 votes)
colinhill
Posted 4 months ago. Direct link to colinhill's post “Alright, I really wanted ...”
Alright, I really wanted to see the difference in bounds graphically, and this is nice. However, I am confused. If (u^3)(du) = 2x(x^2+1) dx, then how are the graphs different? I thought du and dx were both just the bases of an infinite number of rectangles and they both approached zero.

What actually happens when we substitute a variable or when we take the integral with respect to a different variable? I thought they would be equivalent, but now I'm more confused.
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Venkata
  Posted 4 months ago. Direct link to Venkata's post “Good question! I'll try t...”
  Good question! I'll try to put in a simple manner.
  
  So, we have two variables here: x and u. Think of it as two worlds: the (x,f(x)) worlds (where input is x and output is f(x)) and the (u,f(u)) world (where input is u and output is f(u)). Now, the graph will look different in both these worlds. What a u-substitution does is that it creates a map from the x world to the u world (i.e. the substitution we make maps every value of x to a corresponding value of u). As a result, every point is mapped onto a new coordinate system where u = x^2 + 1. This makes our graph into something we can easily find the area under.
  
  Also, a more solid reason as to why the graphs will be different is because of the substitution. See that u = x^2 + 1. If u = x, then the graphs would be the same. But, u = x is a useless substitution, as it just changes the variable and doesn't make anything easier for us.
  
  If you can understand this idea, and you choose to pursue calculus in the future, this visualization helps tremendously when you reach multivariable calculus.
  Comment on Venkata's post “Good question! I'll try t...”
  (4 votes)
Kaylagrey
Posted 2 years 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 2 years 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.
  Button navigates to signup page
  (2 votes)
orel7403
Posted 3 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 3 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.
  Button navigates to signup page
  (1 vote)
LogarithmicLava17
Posted 5 months ago. Direct link to LogarithmicLava17's post “I am extremely confused o...”
I am extremely confused on u-substitution, if the integral was the same thing except going from -1 to 2 instead of 1 to 2, wouldn't u range from 2 to 5? Therefore, you would get the exact same answer even with a different range. In this particular case, it wouldn't matter because the graph of y=2x(x^2+1)^3 is an odd function and the negative area from -1 to 0 would cancel out the positive area from 0 to 1, but in other general cases the u expression might give the same range even though the original range is different. For example, if the u expression has the same output for multiple inputs, then multiple different inputs to the u expression might give the same range especially if the u expression contains a "wavering" function like sin or cos that gives the same output to an infinite number of inputs.
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(1 vote)
Answer
石乐志大师
Posted a year 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 a year 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.
  Button navigates to signup page
  (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?
Button navigates to signup pageComment on gunank312's post “i tried to plot graph of ...”
(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
  Button navigates to signup page
  (1 vote)
caiyresl
Posted 4 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
- Jerry Nilsson
  Posted 4 years ago. Direct link to Jerry Nilsson's post “Yes, after taking the ant...”
  Yes, after taking the antiderivative with respect to 𝑢, we can substitute back to get the antiderivative as a function of 𝑥 and evaluate that for the original boundaries.
  Button navigates to signup page
  (2 votes)

Integral Calculus

Course: Integral Calculus > Unit 1

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