Lessons 13 · 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
13
Antiderivatives and the indefinite integral
Textbook, Part 1: pp. 79–85
GoalMaster the definitions of antiderivative and indefinite integral, write all antiderivatives as F(x) + C and find the one passing through a given point.
New words
antiderivative · boshlang‘ich funksiyaindefinite integral · aniqmas integralintegrand · integral ostidagi funksiyaarbitrary constant C · ixtiyoriy o‘zgarmas C
Explanation
If F′(x) = f(x) on some interval, F(x) is called an antiderivative of f(x) there; it is the inverse of differentiation. For example, if the velocity v(t) is given, the position s(t) is an antiderivative of v. An antiderivative is not unique: if F′(x) = f(x), then (F(x) + C)′ = f(x) as well (the derivative of a constant is 0), and all antiderivatives of f have the form F(x) + C. This family is called the indefinite integral of f(x), written ∫ f(x) dx = F(x) + C (∫ is the integral sign, f(x) the integrand, f(x) dx the expression under the integral). The graphs of F(x) + C are copies of one curve shifted along the Oy axis. To find the antiderivative through a given point, substitute its coordinates into F(x) + C and solve for C. You can always check the result by differentiating: (F(x) + C)′ must equal f(x).
Worked examples
f(x) = 3x²: ∫ 3x² dx = x³ + C, since (x³ + C)′ = 3x². The antiderivative through (2, 10): 8 + C = 10 ⇒ C = 2, F(x) = x³ + 2. Check: F′ = 3x² and F(2) = 10.
A point has velocity v(t) = 6t + 2 (m/s) and s = 0 at t = 0. s(t) = ∫ (6t + 2) dt = 3t² + 2t + C; s(0) = 0 ⇒ C = 0. The distance in 4 s is s(4) = 48 + 8 = 56 m. Also: ∫ cos x dx = sin x + C and ∫ dx/x = ln|x| + C.
Class activity
“The way back”: each pair differentiates a function and hands the result to another pair. They find an antiderivative of that derivative and discuss what determines C (is an extra condition needed?). Then compare with the original function.
Practice
1
Show that F(x) = x⁴ − 2x + 7 is an antiderivative of f(x) = 4x³ − 2.
F′(x) = 4x³ − 2 + 0 = f(x).
2
Write all antiderivatives of f(x) = 5x⁴.
x⁵ + C
3
Find the antiderivative of f(x) = 2x + 1 whose graph passes through (1, 4).
F = x² + x + C; 1 + 1 + C = 4 ⇒ C = 2; F(x) = x² + x + 2.
4
Both x² and x² + 5 are antiderivatives of 2x. Is that a contradiction? Explain.
No contradiction: the derivative of a constant is 0, so both have the same derivative. All antiderivatives are x² + C and their graphs are shifts along Oy.