Week 13 - Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is appropriately named because it establishes a connection between the two branches of calculus: differential calculus and integral calculus.

1. Theorem

  • Suppose f:[a,b]Rf: [a, b] \to \mathbb{R} is continuous. Let FF be the accumulation function, given by F(x)=axf(t)dtF(x) = \int_a^x f(t) dt . Then FF is continuous on [a,b][a, b], differentiable on (a, b), and F(x)=f(x)F'(x) = f(x).

    • To prove: F(x)=F(x+h)F(x)hf(x)hh=f(x)F'(x) = \frac{F(x+h) - F(x)}{h} \approx \frac{f(x) \cdot h}{h} = f(x)
  • Supposef:[a,b]Rf: [a, b] \to \mathbb{R} is continuous, and FF is an antidirevative of ff. Then abf(x)dx=F(b)F(a)\int_a^b f(x) dx = F(b) - F(a)

    • To prove:
      • We say G(x)G(x) is the antidirevative of ff. Then, G(x)=f(x)G'(x) = f(x), and also, we know f(x)=F(x)f(x) = F'(x).
        • where F(x)=axf(t)dtF(x) = \int_a^x f(t) dt
      • In the chapter of antiderivative we have proved that, the antiderivative of ff should be some function plus a constant, then we can get F(x)=G(x)+CF(x) = G(x) + C
        • F(x)F(x) and G(x)G(x) are both the antidirevative functions of ff.
      • Another fact, we know, is: F(a)=aaf(t)dt=0=G(a)+CF(a) = \int_a^a f(t) dt = 0 = G(a) + C, so C=G(a)C = -G(a).
      • Then, we get F(x)=G(x)G(a)F(x) = G(x) - G(a), so axf(t)dt=G(x)G(a)\int_a^x f(t) dt = G(x) - G(a)
      • Let x = b, we get: abf(t)dt=G(b)G(a)=(F(b)C)(F(a)C)=F(b)F(a)\int_a^b f(t) dt = G(b) - G(a) = (F(b) - C) - (F(a) - C) = F(b) - F(a)

1.1. Examples

  • 0πsinxdx\int_0^\pi \sin{x} dx:
    • F(x)=0πsinxdxF(x)=f(x)=sinxF(x)=cosx+C0πsinxdx=F(π)F(0)=cosπ(cos0)=(1)(1)=2\begin{aligned} F(x) &= \int_0^\pi \sin{x} dx \\ F'(x) &= f(x) = \sin{x} \\ F(x) &= -\cos{x} + C\\ \int_0^\pi \sin{x} dx &= F(\pi) - F(0) \\ &= -\cos{\pi} - (-\cos{0}) \\ &= -(-1) - (-1) = 2 \end{aligned}
  • 01x4\int_0^1 x^4:

    • F(x)=01x4F(x)=f(x)=x4F(x)=x55+C01x4=155055=15\begin{aligned} F(x) &= \int_0^1 x^4 \\ F'(x) &= f(x) = x^4 \\ F(x) &= \frac{x^5}{5} + C \\ \int_0^1 x^4 &= \frac{1^5}{5} - \frac{0^5}{5} \\ &= \frac{1}{5} \end{aligned}
  • The area between x\sqrt{x} and x2x^2:

    • 01xx2dx=[x3/23/2x33]01=(2313)0=13\int_0^1 \sqrt{x} - x^2 dx = [\frac{x^{3/2} }{3/2} - \frac{x^3}{3}]_0^1 = (\frac{2}{3} - \frac{1}{3}) - 0 = \frac{1}{3}

  • 0t1x2dx\int_0^t\sqrt{1-x^2}dx

    • Method 1 (use geometrical calculation):

      • Break the area to a circular sector and a triangle.
      • Then, we get:
        • the triangle = t1t22\frac{t \cdot \sqrt{1-t^2} }{2}
        • the circular sector = arcsint2\frac{\arcsin{t} }{2}
      • So 0t1x2dx=t1t22+arcsint2\int_0^t\sqrt{1-x^2}dx = \frac{t \cdot \sqrt{1-t^2} }{2} + \frac{\arcsin{t} }{2}
        • The area of the circular is θ2ππr2\frac{\theta}{2\pi} \cdot \pi r^2, where r=1r = 1, so the area is θ2=arcsint2\frac{\theta}{2} = \frac{\arcsin{t} }{2}
    • Method 2 (Use calculus):

      • Let's say F(x)=x1x22+arcsinx2F(x) = \frac{x \cdot \sqrt{1-x^2} }{2} + \frac{\arcsin{x} }{2}. Then we just need to prove F(x)=1x2F'(x) = \sqrt{1-x^2}
      • // skip

2. The Fundamental Theorem in Physics

  • In physics, v(t)v(t) = velocity at time tt, the displacement can written by abv(t)dt\int_a^b v(t) dt (Distance from time t=0 to t=b).
  • Use Riemann sum, we can get:
    • from t=0 to t=h\text{from } t = 0 \text{ to } t = h, D1=hv(0)D_1 = h \cdot v(0)
    • from t=h to t=2h\text{from } t = h \text{ to } t = 2h, D2=D1+hv(h)D_2 = D_1 + h \cdot v(h)
    • from t=2h to t=3h\text{from } t = 2h \text{ to } t = 3h, D3=D2+hv(2h)D_3 = D_2 + h \cdot v(2h)
    • keep add this down to t=bt=b, and as h goes to zero, the Riemann sum will compute abv(t)dt\int_a^b v(t) dt
  • Summarize this, the accumulation function of velocity is displacement, and the derivative of the displacement is velocity.

3. d/da integral f(x) dx from x = a to x =b

  • ddaabf(x)dx\frac{d}{da} \int_a^bf(x)dx
  • We know that ddbabf(x)dx=f(b)\frac{d}{db} \int_a^bf(x)dx = f(b)
    • The rate of the change of the accumulation function is the functions value
  • Imagine we are calculating the area from a to b. When calculating ddb\frac{d}{db}, we want to know how does that integral change when wiggling b.
  • Compare to dda\frac{d}{da}, we want to know how does the integral change when wiggling a.
  • Let's set the x changes h, So the integral's change should be abf(a+h)dxabf(a)dxhhf(a)h=f(a)\frac{\int_a^bf(a+h)dx - \int_a^bf(a)dx}{h} \approx \frac{-hf(a)}{h} = -f(a)

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