Lessons 25–26 · 2 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
25
Applications of the definite integral
Textbook, Part 2: pp. 3–9
GoalLearn to use the definite integral to find the area between curves, the volume of solids of revolution and the work of a variable force.
New words
area between curves · ikki chiziq orasidagi yuzasolid of revolution · aylanish jismivariable force · o‘zgaruvchan kuchwork done · bajarilgan ish
Explanation
If f₂(x) ≥ f₁(x) on [a, b], the area of the figure between the graphs y = f₁(x), y = f₂(x) and the lines x = a, x = b is S = ∫ₐᵇ (f₂(x) − f₁(x)) dx: from the area under the upper curve we subtract the area under the lower one. The limits of integration are the abscissas of the intersection points, found from f₁(x) = f₂(x). If a curvilinear trapezoid is rotated about the Ox axis, the solid’s volume is V = π ∫ₐᵇ f²(x) dx: we add up the volumes πf²(x)dx of thin discs. For rotation about the Oy axis, V = 2π ∫ₐᵇ x·f(x) dx (a ≥ 0). In physics, if a force F(x) moves a body from x = a to x = b, the work done is A = ∫ₐᵇ F(x) dx; for a spring, Hooke’s law gives F = kx. For a constant force the formula reduces to the familiar A = F · s.
Worked examples
The area between y = x² and y = 2x. From x² = 2x we get x = 0 and x = 2; on [0, 2] we have 2x ≥ x². S = ∫₀² (2x − x²) dx = (x² − x³/3) |₀² = 4 − 8/3 = 4/3.
The arc of y = √x on [0, 4] is rotated about Ox: V = π ∫₀⁴ x dx = π · x²/2 |₀⁴ = 8π. The work done in stretching a spring with k = 200 N/m by 0.3 m: A = ∫₀^0.3 200x dx = 100x² |₀^0.3 = 100 · 0.09 = 9 J.
Class activity
“Area on paper”: on squared paper draw y = x² and y = x + 2, shade the region between them and estimate its area by counting squares. Then compute it with an integral and compare the results.
Practice
1
Find the area between y = x + 2 and y = x².
9/2
2
Find the volume of the solid formed by rotating the arc of y = x on [0, 3] about Ox.
9π
3
A spring is acted on by F = 100x (N). Find the work done in stretching it from 0 to 0.2 m.
2 J
4
Why do we compute the area as ∫(f₂ − f₁)dx and not with f₂ + f₁?
∫f₂dx gives the area under the upper curve, which also contains the part under the lower curve, ∫f₁dx. We want only the figure between the two curves, so we subtract the area under f₁.