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Review practice · Section 1 of 3

Questions 1–6 in this collection

Use horizontal rectangles and the complementary region · Recognise a right-endpoint sum and a cube-root limit · +4 more

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  1. 1

    The region RR is bounded by the x-axis, y-axis, y=1y=1 and y=ln⁡xy=\ln x, x>0x>0. The diagram approximates its area by AA, the total area of nn rectangles of height 1/n1/n, whose widths use the lower y-endpoint of each strip. Show that

    A=1n(1−e1−e1/n).A=\frac1n\left(\frac{1-e}{1-e^{1/n}}\right).

    Another finite region SS is bounded by the x-axis, x=ex=e and y=ln⁡xy=\ln x, x>0x>0. Explain how AA approximates the area of SS, and whether this underestimates or overestimates it, giving a reason. Find the exact volume when SS rotates completely about the y-axis.

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    Horizontal rectangles lie inside R, to the left of x=exp(y), between y=0 and y=1. Each uses its lower y-endpoint width. S lies between that curve and x=e, inside the same enclosing rectangle.
    Horizontal rectangles lie inside R, to the left of x=exp(y), between y=0 and y=1. Each uses its lower y-endpoint width. S lies between that curve and x=e, inside the same enclosing rectangle.
    Horizontal rectangles lie inside R, to the left of x=exp(y), between y=0 and y=1. Each uses its lower y-endpoint width. S lies between that curve and x=e, inside the same enclosing rectangle.

    [2014/TJC/I/10]

    CJC · Review problems · Question 1

  2. 2

    (i) For continuous ff, explain with a sketch why

    lim⁡n→∞1n[f(1n)+f(2n)+⋯+f(nn)]=∫01f(x) dx.\lim_{n\to\infty}\frac1n\left[f\left(\frac1n\right)+f\left(\frac2n\right)+\cdots+f\left(\frac nn\right)\right]=\int_0^1f(x)\,dx.
    marks

    (ii) Hence evaluate lim⁡n→∞1n(13+23+⋯+n3n3)\displaystyle\lim_{n\to\infty}\frac1n\left(\frac{\sqrt[3]1+\sqrt[3]2+\cdots+\sqrt[3]n}{\sqrt[3]n}\right). marks

    [N2015/I/3]

    CJC · Review problems · Question 2

  3. 3

    Find ∫0ax1+6x2 dx\displaystyle\int_0^a x\sqrt{1+6x^2}\,dx.

    CJC · Review problems · Question 3

  4. 4

    Find ∫0πx2sin⁡(nx) dx\displaystyle\int_0^\pi x^2\sin(nx)\,dx, where nn is an even integer. Leave the answer in the form kπ2/nk\pi^2/n and determine kk. marks

    [2018/IJC/I/2]

    CJC · Review problems · Question 4

  5. 5

    (a) The ellipse x2/a2+y2/b2=1x^2/a^2+y^2/b^2=1 has b>a>0b>a>0.

    (i) Show its enclosed area is A=4ba∫0aa2−x2 dxA=\tfrac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx. mark Use x=asin⁡θx=a\sin\theta to find AA in terms of a,b,πa,b,\pi. marks

    (ii) Rotate the enclosed region through π\pi radians about the x-axis to form an ellipsoid. Find its volume. marks

    (b) The curved surface area generated by completely rotating a nonnegative curve y=f(x)y=f(x), a≤x≤ba\le x\le b, about the x-axis is

    S=∫ab2πy1+(dy/dx)2 dx.S=\int_a^b2\pi y\sqrt{1+(dy/dx)^2}\,dx.

    Use the circle x2+y2=r2x^2+y^2=r^2, −r≤x≤r-r\le x\le r, to show the surface area of a sphere is 4πr24\pi r^2. marks

    [2019/ACJC/II/3]

    CJC · Review problems · Question 5

  6. 6

    (a) Find the exact value of ∫12x2ln⁡x dx\int_1^2x^2\ln x\,dx. marks

    (b) Find ∫9/3+2x−x2 dx\int9/\sqrt{3+2x-x^2}\,dx. Hence find the exact constant aa satisfying

    ∫0293+2x−x2 dx=∫−1/2aπ∣1−2x∣ dx.\int_0^2\frac9{\sqrt{3+2x-x^2}}\,dx=\int_{-1/2}^{a}\pi|1-2x|\,dx.
    marks

    [2013/RI/I/12]

    CJC · Review problems · Question 6

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