☰ Contents · Mathematics

The definite integral and the Newton–Leibniz formula

Lessons 15 · 1 lessons · M. A. Mirzaahmedov, Sh. N. Ismailov, A. Q. Amanov (algebra and analysis), B. Q. Haydarov (geometry). Mathematics Grade 11, Parts 1 and 2, 1st edition. ZAMIN NASHR, Tashkent, 2018
15

The definite integral and the Newton–Leibniz formula

Textbook, Part 1: pp. 96–105
GoalUnderstand the area of a curvilinear trapezoid, the definite integral and the Newton–Leibniz formula, and use the basic properties of the definite integral.
New words
curvilinear trapezoid · egri chiziqli trapetsiyadefinite integral · aniq integralNewton–Leibniz formula · Nyuton–Leybnis formulasiintegral sum · integral yig‘indi
Explanation

If f(x) ≥ 0 on [a, b], the figure bounded above by the graph of y = f(x), below by the Ox axis and at the sides by the lines x = a and x = b is called a curvilinear trapezoid. Its area function S(x) satisfies S′(x) = f(x), so S is an antiderivative of f, and the area equals F(b) − F(a) for any antiderivative F of f. This number is the definite integral of f over [a, b]: ∫ₐᵇ f(x) dx = F(b) − F(a) (the Newton–Leibniz formula), written briefly F(x)|ₐᵇ. The definite integral is also the limit of integral sums f(ξₖ)Δxₖ: cut the segment into very thin pieces and add the areas of the rectangles. Properties: ∫ₐᵃ f dx = 0; ∫ₐᵇ f dx = −∫_b^a f dx; ∫ₐᶜ f dx = ∫ₐᵇ f dx + ∫_b^c f dx; if f is even, ∫₋ₐᵃ f dx = 2∫₀ᵃ f dx; if f ≥ 0 the integral is ≥ 0; if f ≤ g then ∫f dx ≤ ∫g dx. The definite integral is a number while the indefinite integral is a family of functions; if f changes sign the integral gives a “signed area” (for example ∫₀^π cos x dx = 0).

Worked examples
The curvilinear trapezoid under y = 2x + 1 on [0, 4]: F(x) = x² + x, S = F(4) − F(0) = 20 − 0 = 20 square units. Check: it is an ordinary trapezoid with bases 1 and 9 and height 4, so its area is (1 + 9)/2 · 4 = 20. Also ∫₀² (3x² + 2x) dx = (x³ + x²)|₀² = (8 + 4) − 0 = 12.
∫₁⁴ √x dx = ((2/3)x^(3/2))|₁⁴ = (2/3)(8 − 1) = 14/3. ∫₀^π sin x dx = (−cos x)|₀^π = 1 + 1 = 2. For an even function: ∫₋₂² x² dx = 2∫₀² x² dx = 2 · 8/3 = 16/3. For the odd function x³: ∫₋₁¹ x³ dx = (x⁴/4)|₋₁¹ = 0.
Class activity

“Counting the area”: on squared paper draw y = x² on [0, 4]. Estimate the area by counting squares (whole and half squares separately) and compare with ∫₀⁴ x² dx = 64/3 ≈ 21.3. How does the result change if the squares are made smaller?

Practice
1
Compute ∫₁³ (4x − 1) dx.
2
Compute ∫₀^(π/2) cos x dx.
3
Find the area of the figure bounded by y = x² + 1, x = 0, x = 3 and the Ox axis.
4
Although ∫₋₁¹ x³ dx = 0, why is the area of the figure between y = x³ and Ox not zero?