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MediumIntegralsCalculus0:46AP Calculus AB, unit 6

The Definite Integral as Area

Find the area under y = x² + 1 from 0 to 2 with an antiderivative and the Fundamental Theorem of Calculus: 14/3, about 4.67 square units.

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On-screen maths

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  1. 1. The question0:00
    -0.50.511.522.5123456
    area under y=x2+1y = x^2 + 1 on [0,2][0, 2]
  2. 2. Antiderivative0:14
    1. ∫02(x2+1) dx\int_0^2 (x^2 + 1)\,dx
    2. F(x)=x33+xF(x) = \frac{x^3}{3} + x
  3. 3. Fundamental Theorem0:24
    1. F(2)−F(0)F(2) - F(0)
    2. =(83+2)−0= \left(\frac{8}{3} + 2\right) - 0
    3. =143≈4.67= \frac{14}{3} \approx 4.67
  4. 4. Recap0:40
    • Find any antiderivative FF.
    • ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a).
    • Above the axis, the integral is the area.

About integrals

Antiderivatives, substitution and the Fundamental Theorem.

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