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𝘶-substitution with definite integrals

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 122x(x2+1)3dx.
We notice that 2x is the derivative of x2+1, so u-substitution applies. Let u=x2+1, then du=2xdx. Now we substitute:
122x(x2+1)3dx=12(u)3du
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=2x(x2+1)3 between x=1 and x=2.
Function y = 2 x left parenthesis x squared + 1 right parenthesis cube is graphed. The x-axis goes from 0 to 3. The graph is a curve. The curve starts in quadrant 2, moves upward away from the x-axis to (2, 500). The region between the curve and the x-axis, between x = 1 and x = 2, is shaded.
Now that we changed the curve to y=u3, why should the limits stay the same?
Functions y = 2 x left parenthesis x squared + 1 right parenthesis cube and y = u cubed are graphed together. The graph of y = u cubed starts in quadrant 2, moves upward away from the x-axis and ends at about (3, 27).
Both y=2x(x2+1)3 and y=u3 are graphed. You can see the areas under the curves between x=1 and x=2 (or u=1 and u=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 u correspond to x2+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:
122x(x2+1)3dx=25(u)3du
Functions y = 2 x left parenthesis x squared + 1 right parenthesis cube and y = u cubed are graphed together. The x-axis goes from negative 1 to 6. Each graph moves upward away from the x-axis. The first function ends at (2, 500). The region between the curve and the x-axis between x = 1 and x = 2 is shaded. The second function ends at about (6, 210). The region between the curve and the x-axis, between x = 1 and x = 5, is shaded. The 2 shaded regions look similar in size.
y=u3 is graphed with the area from u=2 to 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:
25u3du=[u44]25=544244=152.25
Remember: When using u-substitution with definite integrals, we must always account for the limits of integration.
Problem 1
Ella was asked to find 15(2x+1)(x2+x)3dx. This is her work:
Step 1: Let u=x2+x
Step 2: du=(2x+1)dx
Step 3:
15(2x+1)(x2+x)3dx=15u3du
Step 4:
15u3du=[u44]15=544144=156
Is Ella's work correct? If not, what is her mistake?
Choose 1 answer:

Problem 2
1215x2(x37)4dx=?
Choose 1 answer:

Want more practice? Try this exercise.

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  • spunky sam red style avatar for user vanessa hili
    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?
    (2 votes)
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    • female robot grace style avatar for user loumast17
      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.
      (8 votes)
  • leafers seedling style avatar for user colinhill
    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.
    (2 votes)
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    • male robot donald style avatar for user Venkata
      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.
      (6 votes)
  • cacteye green style avatar for user Kaylagrey
    Could someone please explain how

    ∫sec(u)tan(u)du
    =​sec(u)+C

    Tnx!
    (1 vote)
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  • blobby green style avatar for user orel7403
    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
    (1 vote)
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    • sneak peak green style avatar for user Jeff
      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.
      (1 vote)
  • male robot hal style avatar for user LogarithmicLava17
    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.
    (1 vote)
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  • leaf green style avatar for user 石乐志大师
    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?
    (1 vote)
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  • blobby green style avatar for user gunank312
    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?
    (1 vote)
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  • sneak peak green style avatar for user escapingperil
    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?
    (1 vote)
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  • aqualine ultimate style avatar for user caiyresl
    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?
    (1 vote)
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