Q5. Evaluate ∫0π/2 cos2 x dx (NCERT Exercise 7.8, Q12, page 271).
cos2 x = (1 + cos 2x)/2, with anti derivative x/2 + (sin 2x)/4. The value is π/4 + 0 − 0 = π/4.
Q6. Evaluate ∫01 xex2 dx (NCERT Exercise 7.8, Q15, page 271).
An anti derivative is ½ex2 (put t = x2). The value is ½(e − 1).
Q7. For f(x) = 3x2, show that F(x) = x3 and G(x) = x3 + 7 give the same value of ∫12 f(x) dx.
F(2) − F(1) = 8 − 1 = 7 and G(2) − G(1) = 15 − 8 = 7. The 7 added to F cancels in the subtraction.
Q8. Find the area under y = x2 from x = 0 to x = 3, and check it against ∫03 x2 dx.
x2 ≥ 0, so the area is the integral: [x3/3] from 0 to 3 = 9 square units.
Q9. Evaluate ∫−11 x3 dx, and say why the answer is not the area between the curve and the axis.
[x4/4] from −1 to 1 = ¼ − ¼ = 0. The part on [−1, 0] is below the axis and counts negatively; the area of the two pieces together is 2 × ¼ = ½.
Q10. Beyond the NCERT text: write ∫01 x2 dx as the limit of a sum of right-end rectangles, and find the limit.
With n strips of width 1/n, the sum is (1/n)[(1/n)2 + (2/n)2 + … + (n/n)2] = (1/n3) × n(n + 1)(2n + 1)/6. As n grows this tends to 2/6 = 1/3, which is F(1) − F(0) for F(x) = x3/3.