Simulation · Maths · Class 12

The area function

From the lesson The definite integral and the fundamental theorem in Integrals. Change the values and watch what happens.

The idea behind it

NCERT §7.7–7.8

A definite integral ∫ₐᵇ f(x) dx has a single numerical value. a is the lower limit and b the upper limit.
For f(x) ≥ 0 on [a, b], it is the area of the region between y = f(x), the x-axis and the lines x = a and x = b.
Area function: fix the left end at a and let the right end move. A(x) = ∫ₐˣ f(t) dt is the area up to x, so it is itself a function of x.
First fundamental theorem: if f is continuous on [a, b], then A′(x) = f(x). The area grows at a rate equal to the height of the curve at the moving edge.
Second fundamental theorem: if f is continuous on [a, b] and F is any antiderivative of f, then ∫ₐᵇ f(x) dx = F(b) − F(a), written [F(x)]ₐᵇ.
The constant C cancels in F(b) − F(a), so it is left out when evaluating a definite integral.
The integrand must be continuous on the whole closed interval. If it is not, the theorem cannot be applied straight across the bad point.
Worked example: ∫₂³ x² dx = [x³/3] from 2 to 3 = 27/3 − 8/3 = 19/3.
Worked example: x/((x + 1)(x + 2)) = −1/(x + 1) + 2/(x + 2), so ∫₁² x/((x + 1)(x + 2)) dx = [2 log(x + 2) − log(x + 1)] from 1 to 2 = log(32/27).
Worked example: ∫₀^(π/4) sin³2t cos 2t dt = [sin⁴2t / 8] from 0 to π/4 = 1/8.

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