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Worked solution

Equate areas using horizontal and vertical strips

For the curve y=x3y=x^3, the region between the curve, x=1x=1, x=qx=q and the x-axis has the same area as the region between the curve, y=1y=1, y=8y=8 and the y-axis, where q>1q>1. Find qq exactly as a1/ba^{1/b} for integers a,ba,b.

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Curve y=x cubed with two shaded areas: below the curve between x=1 and x=q, and to the left of the curve between y=1 and y=8. Dashed boundaries identify x=1,x=q,y=1,y=8; q is to the right of x=2.
Curve y=x cubed with two shaded areas: below the curve between x=1 and x=q, and to the left of the curve between y=1 and y=8. Dashed boundaries identify x=1,x=q,y=1,y=8; q is to the right of x=2.
Curve y=x cubed with two shaded areas: below the curve between x=1 and x=q, and to the left of the curve between y=1 and y=8. Dashed boundaries identify x=1,x=q,y=1,y=8; q is to the right of x=2.

[2016/MJC/I/2(b)]

CJC · Discussion · Question 6

Answer

q=461/4q=46^{1/4}, with a=46a=46, b=4b=4.

Write the two areas with the correct strip directions

The first area is ∫1qx3dx\int_1^qx^3dx. For horizontal strips, x=y1/3x=y^{1/3}, so the second is ∫18y1/3dy\int_1^8y^{1/3}dy.

How did this part go?

Evaluate and solve for q

(q4−1)/4=[3y4/3/4]18=34(16−1)=45/4(q^4-1)/4=[3y^{4/3}/4]_1^8=\tfrac34(16-1)=45/4. Thus q4=46q^4=46, and since q>1q>1, q=461/4q=46^{1/4}. Hence a=46a=46, b=4b=4.

How did this part go?

Common mistakes

Swap both the function and the limits when changing the strip direction.

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