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∫ {x²+x−3} dx from −1 to 1 (fractional part) | JEE

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Question
The value of ∫−11{x2+x−3} dx\displaystyle\int_{-1}^{1}\{x^{2}+x-3\}\,dx, where { ⋅ }\{\,\cdot\,\} denotes the fractional part of xx, is
A13(1+35)\dfrac{1}{3}\big(1+3\sqrt5\big)
B16(1+35)\dfrac{1}{6}\big(1+3\sqrt5\big)correct
C16(35−1)\dfrac{1}{6}\big(3\sqrt5-1\big)
D13(35−1)\dfrac{1}{3}\big(3\sqrt5-1\big)
Solution
Step 1: Drop the integer constant. Since {y−3}={y}\{y-3\}=\{y\} for any yy, {x2+x−3}={x2+x}\{x^{2}+x-3\}=\{x^{2}+x\}. Write {y}=y−[y]\{y\}=y-[y]:
∫−11{x2+x} dx=∫−11(x2+x) dx−∫−11[x2+x] dx.\int_{-1}^{1}\{x^{2}+x\}\,dx=\int_{-1}^{1}\big(x^{2}+x\big)\,dx-\int_{-1}^{1}\big[x^{2}+x\big]\,dx.
 Step 2: Evaluate the smooth part. The odd term xx integrates to 00:
∫−11(x2+x) dx=∫−11x2 dx=23.\int_{-1}^{1}\big(x^{2}+x\big)\,dx=\int_{-1}^{1}x^{2}\,dx=\dfrac{2}{3}.
 Step 3: Find [x2+x]\big[x^{2}+x\big]. The parabola x2+xx^{2}+x has minimum −14-\dfrac14 at x=−12x=-\dfrac12 and value 00 at x=−1,0x=-1,0, rising to 22 at x=1x=1. On (−1,0)(-1,0): x2+x∈(−14,0)x^{2}+x\in\left(-\tfrac14,0\right), so [x2+x]=−1\big[x^{2}+x\big]=-1. On (0,5−12)\left(0,\tfrac{\sqrt5-1}{2}\right): x2+x∈(0,1)x^{2}+x\in(0,1), so [x2+x]=0\big[x^{2}+x\big]=0 (here x2+x=1x^{2}+x=1 at x=5−12x=\tfrac{\sqrt5-1}{2}). On (5−12,1)\left(\tfrac{\sqrt5-1}{2},1\right): x2+x∈(1,2)x^{2}+x\in(1,2), so [x2+x]=1\big[x^{2}+x\big]=1. Step 4: Integrate the greatest-integer part:
∫−11[x2+x] dx=(−1)(1)+0+1 ⁣(1−5−12)=−1+3−52=1−52.\int_{-1}^{1}\big[x^{2}+x\big]\,dx=(-1)(1)+0+1\!\left(1-\dfrac{\sqrt5-1}{2}\right)=-1+\dfrac{3-\sqrt5}{2}=\dfrac{1-\sqrt5}{2}.
 Therefore
∫−11{x2+x−3} dx=23−1−52=4−3(1−5)6=1+356.\int_{-1}^{1}\{x^{2}+x-3\}\,dx=\dfrac23-\dfrac{1-\sqrt5}{2}=\dfrac{4-3(1-\sqrt5)}{6}=\dfrac{1+3\sqrt5}{6}.
 Correct answer: (2)
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