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Definite integrals, areas and volumes

Practice section 6 of 9

Questions 31–36 in this collection

Practice questions

Discussion practice · 1 of 6

Find a circle–parabola area and half-turn volume

Use x=12cos⁡θx=\sqrt{12}\cos\theta to show that ∫0312−x2 dx=π+33/2\displaystyle\int_0^{\sqrt3}\sqrt{12-x^2}\,dx=\pi+3\sqrt3/2. Region AA is bounded by y=x2y=x^2 and a minor arc of x2+y2=12x^2+y^2=12.

(i) Find its exact area.

(ii) Find the exact volume when AA rotates through π\pi radians about the y-axis.

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Circle x squared plus y squared equals 12 and parabola y=x squared. Region A lies above the parabola and below the upper minor circle arc, between intersections (minus square root 3,3) and (square root 3,3), symmetrically on both sides of the y-axis.
Circle x squared plus y squared equals 12 and parabola y=x squared. Region A lies above the parabola and below the upper minor circle arc, between intersections (minus square root 3,3) and (square root 3,3), symmetrically on both sides of the y-axis.
Circle x squared plus y squared equals 12 and parabola y=x squared. Region A lies above the parabola and below the upper minor circle arc, between intersections (minus square root 3,3) and (square root 3,3), symmetrically on both sides of the y-axis.

[2008/TJC/Promo/13]

CJC · Discussion · Question 7

Solution

Discussion practice · 2 of 6

Compute a composite solid by two methods

Region RR is bounded by y=2xy=2x, y=3/2y=3/2, the x-axis and y=(3x2−1)/x2y=\sqrt{(3x^2-1)/x^2}, as shown. Find the volume when RR rotates through 2π2\pi about the x-axis, correct to three decimal places.

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Region R is above the x-axis until x=1/square root 3, then above the curve square root of (3 minus 1/x squared). Its upper boundary is y=2x until x=3/4 and y=3/2 afterwards, ending at x=2/square root 3.
Region R is above the x-axis until x=1/square root 3, then above the curve square root of (3 minus 1/x squared). Its upper boundary is y=2x until x=3/4 and y=3/2 afterwards, ending at x=2/square root 3.
Region R is above the x-axis until x=1/square root 3, then above the curve square root of (3 minus 1/x squared). Its upper boundary is y=2x until x=3/4 and y=3/2 afterwards, ending at x=2/square root 3.

[2007/HCI/I/11(modified)]

CJC · Discussion · Question 8

Solution

Discussion practice · 3 of 6

Translate a rotation axis before integrating

Curve CC is y=3x/(4−x)y=\sqrt{3x/(4-x)}, 0≤x<40\le x<4. Region RR is bounded by CC, y=3y=3 and the y-axis. It rotates through 2π2\pi about y=3y=3. Write the equation after translating CC three units down, and hence find the solid volume.

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Curve y=square root of 3x/(4 minus x) starts at the origin and meets the horizontal rotation axis y=3 at (3,3); x=4 is a vertical asymptote. Region R is shaded between y=3, the curve and the y-axis from x=0 to x=3.
Curve y=square root of 3x/(4 minus x) starts at the origin and meets the horizontal rotation axis y=3 at (3,3); x=4 is a vertical asymptote. Region R is shaded between y=3, the curve and the y-axis from x=0 to x=3.
Curve y=square root of 3x/(4 minus x) starts at the origin and meets the horizontal rotation axis y=3 at (3,3); x=4 is a vertical asymptote. Region R is shaded between y=3, the curve and the y-axis from x=0 to x=3.

[2016/RVHS/II/4(modified)]

CJC · Discussion · Question 9

Solution

Review practice · 4 of 6

Use horizontal rectangles and the complementary region

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

Solution

Review practice · 5 of 6

Recognise a right-endpoint sum and a cube-root limit

(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.

[2]

(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). [3]

[N2015/I/3]

CJC · Review problems · Question 2

Solution

Review practice · 6 of 6

Match the inner derivative in a radical integral

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

CJC · Review problems · Question 3

Solution

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