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If A is the area of the region enclosed between the curves y = x*ln(x) and y = 2x - 2x², find the value of 12A.

1. 1
2. 2
3. 3
4. 4

Correct answer: 1

Solution

Curves: y1 = x*ln(x), y2 = 2x-2x². Both pass through (0,0) (using limit x*ln x -> 0 as x->0+) and (1,0). On (0,1): at x=1/2: y1=(1/2)*ln(1/2)=(1/2)*(-ln2)≈-0.347; y2=2*(1/2)-2*(1/4)=1-0.5=0.5. So y2 > y1 on (0,1). A = integral₀¹ (y2-y1) dx = integral₀¹ (2x-2x²-x*ln x) dx. Integral of 2x dx = x²; from 0 to 1 = 1. Integral of 2x² dx = 2x³/3; from 0 to 1 = 2/3. Integral of x*ln x dx: by parts, u=ln x, dv=x dx; du=dx/x, v=x²/2. = [x²*ln x/2]₀¹ - integral x/2 dx = 0 - [x²/4]₀¹ = -1/4. So integral of x*ln x from 0 to 1 = -1/4. A = 1 - 2/3 - (-1/4) = 1 - 2/3 + 1/4 = 12/12 - 8/12 + 3/12 = 7/12. 12A = 7. But 7 is not among options 1-4. Let me recheck: at x=1/2, y2=0.5 > y1=-0.347. Area calculation: 1 - 2/3 + 1/4 = 12/12 - 8/12 + 3/12 = 7/12. 12A=7. Hmm options show 1,2,3,4. This doesn't match. Perhaps the answer is displayed as '1' in the options meaning answer choice 1 corresponds to 7? Or there's a typo. 12A=7 is the mathematical result.

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