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Chapter Analysis
Intermediate8 pages • EnglishQuick Summary
The chapter 'Application of Integrals' focuses on using integrals to find areas under curves, between lines, and arcs of various conic sections like circles, parabolas, and ellipses. It builds on the concept of definite integrals, explaining how they can be employed to calculate areas, emphasizing both symbolic and practical approaches. Through examples, it demonstrates calculating areas in different geometric scenarios, considering both vertical and horizontal strips.
Key Topics
- •Area under the curve
- •Area between curves
- •Applications of definite integrals
- •Areas using vertical and horizontal strips
- •Integral Calculus in conic sections
- •Definite integrals as limit of sums
Learning Objectives
- ✓Understand how to apply integrals to calculate areas under curves.
- ✓Learn the concept of definite integrals as limits of sums.
- ✓Calculate the areas between various geometric shapes including circles and ellipses.
- ✓Apply integration to practical geometric problems.
Questions in Chapter
Find the area of the region bounded by the ellipse x^2/16 + y^2/9 = 1.
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Find the area of the region bounded by the ellipse x^2/4 + y^2/9 = 1.
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Area lying in the first quadrant and bounded by the circle x^2 + y^2 = 4 and the lines x = 0 and x = 2 is (A) π (B) π/2 (C) 3π/4 (D) π/4
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Area of the region bounded by the curve y^2 = 4x, y-axis and the line y = 3 is (A) 2 (B) 9/4 (C) 3 (D) 9/2
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Find the area bounded by the curve y = sin x between x = 0 and x = 2π.
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Additional Practice Questions
Calculate the area enclosed by the parabola y^2 = 4ax and the line x = a.
mediumAnswer: The area enclosed by the parabola y^2 = 4ax and x = a is obtained by integrating √(4ax) from x = 0 to x = a. The computed area is (4/3)a^2.
Determine the area between the curves y = x^2 and y = x over the interval [0,1].
easyAnswer: The area between y = x^2 and y = x over [0,1] is found by integrating (x-x^2) from 0 to 1. The computed area is 1/6.
Find the area of the region lying between the lines y = x + 2 and y = -x + 2.
mediumAnswer: The lines intersect at (0,2) and (2,0). Integration of (x + 2) - (-x + 2) from 0 to 2 gives an area of 4.
Compute the area under the sine curve from x = 0 to x = π.
easyAnswer: The area under one period of sine from 0 to π is ∫sin(x) dx from 0 to π which results in 2 units.
Evaluate the integral of the function f(x) = x^3 from x = -1 to x = 1 and interpret the result geometrically.
easyAnswer: The integral of f(x) = x^3 from -1 to 1 is zero due to symmetry. Geometrically, this means the areas above and below the x-axis cancel each other.