Q5. Find the area between the parabola y = 9 − x2 and the x-axis.
The parabola meets the axis at x = ±3 and is above it between them. ∫−33 (9 − x2) dx = 54 − 18 = 36.
Q6. Find the area between y = x2 and y = 8 − x2.
They meet where x2 = 4, at ±2; 8 − x2 is on top. ∫−22 (8 − 2x2) dx = 32 − 32/3 = 64/3.
Q7. Find the area of the region bounded by y = |x| and y = 2.
The lines meet at (−2, 2) and (2, 2). The region is a triangle with base 4 and height 2, area 4; by integration, ∫−22 (2 − |x|) dx = 2 × (4 − 2) = 4.
Q8. Find the area enclosed between the parabola y2 = 4x and the line y = 2x.
They meet where 4x2 = 4x, at x = 0 and 1. The parabola’s upper branch 2√x is on top: ∫01 (2√x − 2x) dx = 4/3 − 1 = 1/3.
Q9. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 16.
The line meets the circle at (2√2, 2√2), making an angle of π/4 with the x-axis. The region is a sector of radius 4 with angle π/4, so its area is ½ × 16 × π/4 = 2π.
Q10. Find the area between y = ex, y = e−x and the line x = 1.
The curves meet at x = 0; for 0 < x ≤ 1, ex is on top. ∫01 (ex − e−x) dx = (e − 1) + (e−1 − 1) = e + 1/e − 2.
Q11. Beyond the NCERT text (JEE): find the area of the region {(x, y) : 0 ≤ y ≤ x2 + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}.
y lies under both curves, so the top is the lower one: x2 + 1 on [0, 1], where x2 ≤ x, and x + 1 on [1, 2]. The area is ∫01 (x2 + 1) dx + ∫12 (x + 1) dx = 4/3 + 5/2 = 23/6.