Application of Integrals
Karnataka Board · Class 12 · Mathematics
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Find the area bounded by the curve y = x², the x-axis, and the lines x = 1 and x = 3.
The area bounded by y = 4 - x² and the x-axis is:
Find the area between the curves y = x² and y = 4x - x².
Which method is most appropriate for finding the area bounded by x = y² - 1 and x = 3?
Sample Questions
Which of the following integrals correctly represent the area bounded by y = sin x, x-axis, from x = 0 to x = π?
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∫₀^π sin x dx, ∫₀^π |sin x| dx, 2∫₀^(π/2) sin x dx
Since sin x ≥ 0 on [0,π], the area is ∫₀^π sin x dx = 2. The absolute value notation is also correct. Due to symmetry, 2∫₀^(π/2) sin x dx also gives the same result.
Calculate the area bounded by y = √x, x-axis, and the line x = 4.
True or False: The area bounded by y = x³, x-axis, from x = -1 to x = 1 is zero.
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True
∫₋₁¹ x³ dx = [x⁴/4]₋₁¹ = 1/4 - 1/4 = 0. This is because x³ is an odd function, so the areas above and below the x-axis cancel out over symmetric intervals.
The area of the region bounded by x = y² and x = 4 is:
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Sources & Official References
- Karnataka SSLC — kseeb.kar.nic.in
- Dept of Pre-University Education, Karnataka
- National Education Policy 2020 — education.gov.in
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