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Fundamental theorem of calculus review

Review your knowledge of the fundamental theorem of calculus and use it to solve problems.

What is the fundamental theorem of calculus?

The theorem has two versions.

a) start fraction, d, divided by, d, x, end fraction, integral, start subscript, a, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t, equals, f, left parenthesis, x, right parenthesis

We start with a continuous function f and we define a new function for the area under the curve y, equals, f, left parenthesis, t, right parenthesis:
F, left parenthesis, x, right parenthesis, equals, integral, start subscript, a, end subscript, start superscript, x, end superscript, f, left parenthesis, t, right parenthesis, d, t
What this version of the theorem says is that the derivative of F is f. In other words, F is an antiderivative of f. Thus, the theorem relates differential and integral calculus, and tells us how we can find the area under a curve using antidifferentiation.

b) integral, start subscript, a, end subscript, start superscript, b, end superscript, f, left parenthesis, x, right parenthesis, d, x, equals, F, left parenthesis, b, right parenthesis, minus, F, left parenthesis, a, right parenthesis

This version gives more direct instructions to finding the area under the curve y, equals, f, left parenthesis, x, right parenthesis between x, equals, a and x, equals, b. Simply find an antiderivative F and take F, left parenthesis, b, right parenthesis, minus, F, left parenthesis, a, right parenthesis.
Want to learn more about the fundamental theorem of calculus? Check out this video.

Practice set 1: Applying the theorem

Problem 1.1
g, left parenthesis, x, right parenthesis, equals, integral, start subscript, 1, end subscript, start superscript, x, end superscript, square root of, 2, t, plus, 7, end square root, d, t
g, prime, left parenthesis, 9, right parenthesis, equals
  • Your answer should be
  • an integer, like 6
  • an exact decimal, like 0, point, 75
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4

Want to try more problems like this? Check out this exercise.

Practice set 2: Applying the theorem with chain rule

We can use the theorem in more hairy situations. Let's find, for example, the expression for start fraction, d, divided by, d, x, end fraction, integral, start subscript, 0, end subscript, start superscript, x, cubed, end superscript, sine, left parenthesis, t, right parenthesis, d, t. Note that the interval is between 0 and x, cubed, not x.
To help us, we define F, left parenthesis, x, right parenthesis, equals, integral, start subscript, 0, end subscript, start superscript, x, end superscript, sine, left parenthesis, t, right parenthesis, d, t. According to the fundamental theorem of calculus, F, prime, left parenthesis, x, right parenthesis, equals, sine, left parenthesis, x, right parenthesis.
It follows from our definition that integral, start subscript, 0, end subscript, start superscript, x, cubed, end superscript, sine, left parenthesis, t, right parenthesis, d, t is F, left parenthesis, x, cubed, right parenthesis, which means that start fraction, d, divided by, d, x, end fraction, integral, start subscript, 0, end subscript, start superscript, x, cubed, end superscript, sine, left parenthesis, t, right parenthesis, d, t is start fraction, d, divided by, d, x, end fraction, F, left parenthesis, x, cubed, right parenthesis. Now we can use the chain rule:
=ddx0x3sin(t)dt=ddxF(x3)=F(x3)ddx(x3)=sin(x3)3x2\begin{aligned} &\phantom{=}\dfrac{d}{dx}\displaystyle \int_{0}^{x^3}\sin(t) \, dt \\\\ &=\dfrac{d}{dx}F(x^3) \\\\ &=F'(x^3)\cdot\dfrac{d}{dx}(x^3) \\\\ &=\sin(x^3)\cdot 3x^2 \end{aligned}
Problem 2.1
F, left parenthesis, x, right parenthesis, equals, integral, start subscript, 0, end subscript, start superscript, x, start superscript, 4, end superscript, end superscript, cosine, left parenthesis, t, right parenthesis, d, t
F, prime, left parenthesis, x, right parenthesis, equals

Want to try more problems like this? Check out this exercise.

Want to join the conversation?

  • leaf green style avatar for user allan1.davis
    I can't convince myself why these expressions should be treated as more than one function, thus requiring the use of the chain rule.
    (17 votes)
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  • leaf blue style avatar for user Victor Mayland Nielsen
    On the integral sign you go from one point to another. I don't get why this point can be x or something with x. X is not a point on the x axis. It just doesn't make sense. From my understanding, the two points on an integral are two places on the x-axis.
    (2 votes)
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  • blobby green style avatar for user John Vanderwalls
    I'm having trouble with this equality written in the text: d/dx F(x^3)= F'(x^3).(x^3)'

    What's the difference between d/dx and ' ?
    (3 votes)
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  • blobby green style avatar for user bgautam0707
    In Practice set 2, we had to find F(x^3), so we took help of F(x).
    but, problem 2.1 says that F(x) is equal to integral from 0 to x^4, i think it should be F(x^4).
    please help.
    (3 votes)
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  • leaf green style avatar for user beilingliu77
    I still don't understand, but, so the antiderivative F is same as definite integral. If we find the area under the curve, which is definite integral, we can say that, this function is equal to anti-derivative?
    (1 vote)
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    • leafers ultimate style avatar for user Gavin Yu
      No, the antiderivative is the same as the indefinite integral, which doesn't represent area. An antiderivative is only a function such that its derivative is your original function.

      The definite integral represents the area under a function between two points. Antiderivatives can help us find definite integrals.
      (3 votes)
  • duskpin seed style avatar for user ximena.tamariz
    I think I understand the theorem when the lower limit is zero and the upper limit is x, but what do you do when the lower limit is 2 and the upper limit is x. A specific problem I have is: d/dx(∫(upper limit x, lower limit 2)sin(t^4)dt)
    (2 votes)
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  • aqualine ultimate style avatar for user Evana Holevas
    What is the difference between the FTC and the Antiderivative Method? I'm working on a problem that wants me to use both formulas and it says the functions will be different however I'm having trouble remembering. Would someone mind explaining?
    (2 votes)
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  • leaf green style avatar for user beilingliu77
    Using, problem 2.1 as example
    when using the chain rule, why does F'(x) stay the same as COS(X^4)*4X^3?
    I thought the chain rule should be
    f(x)=cos(x)
    f'(x)=-sin(x)
    g(x)=x^4
    g'(x)=4x^3

    apply chain rule here, should be f'(g(x))*g'(x)
    which should be f'(x^4)*4x^3
    -sin(x^4)*4x^3

    or am I confused with something? Thanks for the help
    (1 vote)
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    • leaf green style avatar for user Mateusz Jastrzębski
      Your conclusions are alright but you're not solving for what's being asked.

      The definite integral equals F(x)=Integral(f(t)) from 0 to x^4. Now, if you take the derivative of this integral you get f(x^4) times d/dx(x^4). You don't differentiate the f(t) because it is in fact your original function before integration.

      Fundamental Theorem of Calculus is tricky to understand but once you know it by heart it'll never leave you. If you're struggling to get a good grasp on this fundamental concept try this: check out this video - https://www.khanacademy.org/math/ap-calculus-ab/ab-antiderivatives-ftc/ab-fundamental-theorem-of-calc/v/fundamental-theorem-of-calculus - and consider the following: it is a very subtle and simple idea:

      On a diagram, there's a yellow line with an x label. What is this yellow line? It's f(x). What does its value represent? It represents the rate of change of F around that point x and naturally it means it is the derivative of the integral F. The slope of F at point x is equal to f(x)
      (1 vote)
  • starky sapling style avatar for user 20leunge
    So is the derivative the inverse of the integral?
    (1 vote)
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  • mr pants teal style avatar for user ddan281847
    I don't really understand how a derivative of this is just the function of the equation that was used. But the difference is if x matters since the part being cut is constantly being increase as x increases and the functions derivative is the same as the function being measured?
    (1 vote)
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    • duskpin tree style avatar for user ThePencilThief
      The reason why the derivative is the function of the eqution is because the FTC is defined in such a way, F'(x)=f(x), two separate functions linked in that way.
      An integral can also be stated as an anti-derivative, so the opposite of a derivative. If you take the derivative of an integral, it is like taking the square root of x squared, they cancel out, and you are left with x. More can be seen as we take the limits of integrals as x approaches infinite, but we are not yet there in these videos. Take time to really get a good foundation in the FTC because it is VITAL to calculus understanding. I hope I answered your question properly, have a good day!
      (1 vote)