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Review practice · Section 2 of 3

Questions 7–12 in this collection

Reverse reciprocal-substitution bounds and reuse an integral · Combine inverse-sine integration with two rotation radii · +4 more

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  1. 7

    (a)(i) For a>1a>1, use u=1/xu=1/x to show ∫1adx/(1+x2)=∫1/a1du/(1+u2)\int_1^a dx/(1+x^2)=\int_{1/a}^1du/(1+u^2). marks

    (ii) Hence show tan⁡−1a+tan⁡−1(1/a)=k\tan^{-1}a+\tan^{-1}(1/a)=k, finding kk. marks

    (b)(i) Show ∫01x3/(1+x2) dx=1/2−(1/2)ln⁡2\int_0^1x^3/(1+x^2)\,dx=1/2-(1/2)\ln2. marks

    (ii) Without using a calculator, evaluate ∫01x2tan⁡−1x dx\int_0^1x^2\tan^{-1}x\,dx. marks

    [2013/JJC/I/12]

    CJC · Review problems · Question 7

  2. 8

    (a) Find ∫sin⁡−1(3x) dx\int\sin^{-1}(3x)\,dx. marks

    (b) The curves C1,C2C_1,C_2 are y=1−(3/π)x2y=1-(\sqrt3/\pi)x^2 and y=13cos⁡xy=\tfrac13\cos x. Rotate the first-quadrant region bounded by them and the y-axis through 360∘360^\circ about the y-axis. Find the volume to 2 decimal places. marks

    [2018/HCI/Promo/3]

    CJC · Review problems · Question 8

  3. 9

    The diagram shows y=sin⁡−1(x/2)y=\sin^{-1}(x/2) for 0≤x≤20\le x\le2.

    (i) Find the area of R1R_1, bounded by the curve, y=π/12y=\pi/12, y=π/6y=\pi/6 and the y-axis.

    (ii) Find the volume when R1R_1 rotates through 2π2\pi about the x-axis.

    (iii) Without using a calculator, find the exact area of R2R_2, bounded by the curve, x=1x=1, x=3x=\sqrt3 and the x-axis.

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    Given curve y=arcsin(x/2), 0≤x≤2. R1 is to the left of the curve between y=pi/12 and pi/6. R2 lies below the curve between x=1 and sqrt(3). Both named regions and boundary lines are shown.
    Given curve y=arcsin(x/2), 0≤x≤2. R1 is to the left of the curve between y=pi/12 and pi/6. R2 lies below the curve between x=1 and sqrt(3). Both named regions and boundary lines are shown.
    Given curve y=arcsin(x/2), 0≤x≤2. R1 is to the left of the curve between y=pi/12 and pi/6. R2 lies below the curve between x=1 and sqrt(3). Both named regions and boundary lines are shown.

    [2015/DHS/I/9]

    CJC · Review problems · Question 9

  4. 10

    The curve CC is y=xcos⁡2xy=x\cos2x, 0≤x≤π0\le x\le\pi.

    (i) Find its exact x-axis intercepts. marks

    (ii) Sketch CC, stating all x- and y-axis intercept coordinates. marks

    (iii) Find the exact value of ∫π/4π∣xcos⁡2x∣ dx\int_{\pi/4}^\pi|x\cos2x|\,dx. marks

    (iv) Find the volume when the region bounded by CC, the x-axis and x=πx=\pi rotates completely about the x-axis. marks

    [2014/IJC/I/11]

    CJC · Review problems · Question 10

  5. 11

    Sketch y=exy=e^{\sqrt x}, x≥0x\ge0. mark Rotate the region bounded by this curve, x=4x=4 and both axes completely about the x-axis. Use t=xt=\sqrt x to show the volume is 2π∫02te2t dt2\pi\int_0^2te^{2t}\,dt, and evaluate it exactly. marks

    [2014/YJC/II/3(c)]

    CJC · Review problems · Question 11

  6. 12

    The diagram shows y2=6−xy^2=6-x and y=−xy=-x.

    (i) State their intersection coordinates.

    (ii) Find the exact area of the finite region between the two graphs.

    (iii) The shaded region RR is bounded by the y-axis, the line and the lower part of the parabola. Find the volume when RR rotates through 2π2\pi about the y-axis, to 2 decimal places.

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    Parabola x=6−y² and line x=−y intersect at (−3,3) and (2,−2). R is shaded between the y-axis and the nearer positive-x boundary, from y=−sqrt(6) to zero; it changes boundary at (2,−2).
    Parabola x=6−y² and line x=−y intersect at (−3,3) and (2,−2). R is shaded between the y-axis and the nearer positive-x boundary, from y=−sqrt(6) to zero; it changes boundary at (2,−2).
    Parabola x=6−y² and line x=−y intersect at (−3,3) and (2,−2). R is shaded between the y-axis and the nearer positive-x boundary, from y=−sqrt(6) to zero; it changes boundary at (2,−2).

    [2008/RJC/Promo/9(modified)]

    CJC · Review problems · Question 12

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