2020HSCQ10Which of the following is equal to ∫02af(x) dx\displaystyle\int_0^{2a}f(x)\,dx∫02af(x)dx? A. ∫0a[f(x)−f(2a−x)] dx\displaystyle\int_0^a[f(x)-f(2a-x)]\,dx∫0a[f(x)−f(2a−x)]dx B. ∫0a[f(x)+f(2a−x)] dx\displaystyle\int_0^a[f(x)+f(2a-x)]\,dx∫0a[f(x)+f(2a−x)]dx C. 2∫0af(x−a) dx\displaystyle2\int_0^a f(x-a)\,dx2∫0af(x−a)dx D. ∫0a12f(2x) dx\displaystyle\int_0^a\frac12f(2x)\,dx∫0a21f(2x)dx1 markMultiple choice
2020HSCQ11 (b)Use integration by parts to evaluate ∫1exlnx dx.\int_1^e x\ln x\,dx.∫1exlnxdx.3 marksShort answer
2020HSCQ13 (d) (iii)Hence, or otherwise, find ∫0π/2cos4θ dθ.\int_0^{\pi/2}\cos^4\theta\,d\theta.∫0π/2cos4θdθ.2 marksShort answer
2020HSCQ16 (b) (i)Let In=∫0π/2sin2n+1(2θ) dθ,n=0,1,….I_n=\int_0^{\pi/2}\sin^{2n+1}(2\theta)\,d\theta,\qquad n=0,1,\ldots.In=∫0π/2sin2n+1(2θ)dθ,n=0,1,…. (i) Prove that In=2n2n+1In−1,n≥1.I_n=\frac{2n}{2n+1}I_{n-1},\qquad n\geq1.In=2n+12n3 marksShort answer
2020HSCQ16 (b) (iv)Prove that (2nn!)2≤(2n+1)!.\bigl(2^n n!\bigr)^2\leq(2n+1)!.(2nn!)2≤(2n+1)!.2 marksShort answer
2020HSCQ16 (b) (ii)Using the result of part (i), deduce that In=22n(n!)2(2n+1)!.I_n=\frac{2^{2n}(n!)^2}{(2n+1)!}.In=(2n+1)!22n(n!)2.3 marksShort answer
2020HSCQ16 (b) (iii)Let Jn=∫01xn(1−x)n dx,n=0,1,2,….J_n=\int_0^1x^n(1-x)^n\,dx,\qquad n=0,1,2,\ldots.Jn=∫01xn(1−x)ndx,n=0,1,2,…. Using the result of part (ii), or otherwise, show that Jn=(n!)2(2n+1)!.J_n=\frac{(n!)^2}{(2n+1)!}.Jn=(2n+1)!(n!3 marksShort answer