[문제 3-1]
+4점 · 정적분을 쓸 수 있도록 분자를 변형하면 ( 3 + k n ) ln ( 3 n + k ) − ( 2 + k n ) ln ( 2 n + k ) − ln n = ( 3 + k n ) ln n ( 3 + k n ) − ( 2 + k n ) ln n ( 2 + k n ) − ln n = ( 3 + k n ) ln ( 3 + k n ) − ( 2 + k n ) ln ( 2 + k n ) + ( 3 + k n ) ln n − ( 2 + k n ) ln n − ln n = ( 3 + k n ) ln ( 3 + k n ) − ( 2 + k n ) ln ( 2 + k n ) \displaystyle \begin{aligned}&\left(3+\frac kn\right)\ln(3n+k)-\left(2+\frac kn\right)\ln(2n+k)-\ln n\\&=\left(3+\frac kn\right)\ln n\left(3+\frac kn\right)-\left(2+\frac kn\right)\ln n\left(2+\frac kn\right)-\ln n\\&=\left(3+\frac kn\right)\ln\left(3+\frac kn\right)-\left(2+\frac kn\right)\ln\left(2+\frac kn\right)+\left(3+\frac kn\right)\ln n-\left(2+\frac kn\right)\ln n-\ln n\\&=\left(3+\frac kn\right)\ln\left(3+\frac kn\right)-\left(2+\frac kn\right)\ln\left(2+\frac kn\right)\end{aligned} ( 3 + n k ) ln ( 3 n + k ) − ( 2 + n k ) ln ( 2 n + k ) − ln n = ( 3 + n k ) ln n ( 3 + n k ) − ( 2 + n k ) ln n ( 2 + n k ) − ln n = ( 3 + n k ) ln ( 3 + n k ) − ( 2 + n k ) ln ( 2 + n k ) + ( 3 + n k ) ln n − ( 2 + n k ) ln n − ln n = ( 3 + n k ) ln ( 3 + n k ) − ( 2 + n k ) ln ( 2 + n k ) 이므로 주어진 극한값은 다음 정적분이다.
+2점 · 맞는 정적분 형태로 쓰면 lim n → ∞ ∑ k = 1 n 1 n ( 3 + k n ) ln ( 3 + k n ) − ( 2 + k n ) ln ( 2 + k n ) ( 2 + k n ) ( 3 + k n ) ln ( 2 + k n ) ln ( 3 + k n ) = ∫ 0 1 ( 3 + x ) ln ( 3 + x ) − ( 2 + x ) ln ( 2 + x ) ( 2 + x ) ( 3 + x ) ln ( 2 + x ) ln ( 3 + x ) d x = ∫ 0 1 1 ( 2 + x ) ln ( 2 + x ) d x − ∫ 0 1 1 ( 3 + x ) ln ( 3 + x ) d x \displaystyle \begin{aligned}&\lim_{n\to\infty}\sum_{k=1}^n\frac1n\frac{\left(3+\frac kn\right)\ln\left(3+\frac kn\right)-\left(2+\frac kn\right)\ln\left(2+\frac kn\right)}{\left(2+\frac kn\right)\left(3+\frac kn\right)\ln\left(2+\frac kn\right)\ln\left(3+\frac kn\right)}\\&=\int_0^1\frac{(3+x)\ln(3+x)-(2+x)\ln(2+x)}{(2+x)(3+x)\ln(2+x)\ln(3+x)}dx\\&=\int_0^1\frac1{(2+x)\ln(2+x)}dx-\int_0^1\frac1{(3+x)\ln(3+x)}dx\end{aligned} n → ∞ lim k = 1 ∑ n n 1 ( 2 + n k ) ( 3 + n k ) ln ( 2 + n k ) ln ( 3 + n k ) ( 3 + n k ) ln ( 3 + n k ) − ( 2 + n k ) ln ( 2 + n k ) = ∫ 0 1 ( 2 + x ) ( 3 + x ) ln ( 2 + x ) ln ( 3 + x ) ( 3 + x ) ln ( 3 + x ) − ( 2 + x ) ln ( 2 + x ) d x = ∫ 0 1 ( 2 + x ) ln ( 2 + x ) 1 d x − ∫ 0 1 ( 3 + x ) ln ( 3 + x ) 1 d x
+4점 · 2 ln ( ln 3 ) − ln ( ln 2 ) − ln ( ln 4 ) \displaystyle 2\ln(\ln3)-\ln(\ln2)-\ln(\ln4) 2 ln ( ln 3 ) − ln ( ln 2 ) − ln ( ln 4 ) y = ln ( 2 + x ) \displaystyle y=\ln(2+x) y = ln ( 2 + x ) y = ln ( 3 + x ) \displaystyle y=\ln(3+x) y = ln ( 3 + x ) 2 ln ( ln 3 ) − ln ( ln 2 ) − ln ( ln 4 ) \displaystyle 2\ln(\ln3)-\ln(\ln2)-\ln(\ln4) 2 ln ( ln 3 ) − ln ( ln 2 ) − ln ( ln 4 ) 2 ln ( ln 3 ) − 2 ln ( ln 2 ) − ln 2 \displaystyle 2\ln(\ln3)-2\ln(\ln2)-\ln2 2 ln ( ln 3 ) − 2 ln ( ln 2 ) − ln 2 ln ( ( ln 3 ) 2 2 ( ln 2 ) 2 ) \displaystyle \ln\left(\frac{(\ln3)^2}{2(\ln2)^2}\right) ln ( 2 ( ln 2 ) 2 ( ln 3 ) 2 )
[문제 3-2]
+3점 · f ( x ) = ∣ x ∣ + 2 ( x − t ) 2 + 3 t \displaystyle f(x)=|x|+2\sqrt{(x-t)^2+3t} f ( x ) = ∣ x ∣ + 2 ( x − t ) 2 + 3 t f ( x ) = ∣ x ∣ + 2 ( x − t ) 2 + 3 t \displaystyle f(x)=|x|+2\sqrt{(x-t)^2+3t} f ( x ) = ∣ x ∣ + 2 ( x − t ) 2 + 3 t x ≥ 0 \displaystyle x\ge0 x ≥ 0
f ′ ( x ) = 1 + 2 ( x − t ) ( x − t ) 2 + 3 t \displaystyle f^{\prime}(x)=1+\frac{2(x-t)}{\sqrt{(x-t)^2+3t}} f ′ ( x ) = 1 + ( x − t ) 2 + 3 t 2 ( x − t ) x ≥ t \displaystyle x\ge t x ≥ t f ′ ( x ) ≥ 0 \displaystyle f^{\prime}(x)\ge0 f ′ ( x ) ≥ 0 0 ≤ x ≤ t \displaystyle 0\le x\le t 0 ≤ x ≤ t f ′ ( x ) = 0 \displaystyle f^{\prime}(x)=0 f ′ ( x ) = 0 x = t − t \displaystyle x=t-\sqrt t x = t − t
+3점 · 0 ≤ t ≤ 1 \displaystyle 0\le t\le1 0 ≤ t ≤ 1 m ( t ) = 2 t 2 + 3 t \displaystyle m(t)=2\sqrt{t^2+3t} m ( t ) = 2 t 2 + 3 t 0 ≤ t ≤ 1 \displaystyle 0\le t\le1 0 ≤ t ≤ 1 t − t ≤ 0 \displaystyle t-\sqrt t\le0 t − t ≤ 0 x = 0 \displaystyle x=0 x = 0 m ( t ) = 2 t 2 + 3 t \displaystyle m(t)=2\sqrt{t^2+3t} m ( t ) = 2 t 2 + 3 t
+3점 · t ≥ 1 \displaystyle t\ge1 t ≥ 1 m ( t ) = t + 3 t \displaystyle m(t)=t+3\sqrt t m ( t ) = t + 3 t t ≥ 1 \displaystyle t\ge1 t ≥ 1 0 ≤ t − t ≤ t \displaystyle 0\le t-\sqrt t\le t 0 ≤ t − t ≤ t x = t − t \displaystyle x=t-\sqrt t x = t − t m ( t ) = t + 3 t \displaystyle m(t)=t+3\sqrt t m ( t ) = t + 3 t
구하는 적분은 다음과 같다. ∫ 1 2 4 ln ( m ( t ) ) d t = ∫ 1 2 1 ln ( 2 t 2 + 3 t ) d t + ∫ 1 4 ln ( t + 3 t ) d t \displaystyle \int_{\frac12}^4\ln(m(t))dt=\int_{\frac12}^1\ln(2\sqrt{t^2+3t})dt+\int_1^4\ln(t+3\sqrt t)dt ∫ 2 1 4 ln ( m ( t )) d t = ∫ 2 1 1 ln ( 2 t 2 + 3 t ) d t + ∫ 1 4 ln ( t + 3 t ) d t
+3점 · 13 2 ln 2 − 7 4 ln 7 − 1 2 \displaystyle \frac{13}2\ln2-\frac74\ln7-\frac12 2 13 ln 2 − 4 7 ln 7 − 2 1 ∫ 1 2 1 ln ( 2 t 2 + 3 t ) d t = 1 2 ln 2 + ∫ 1 2 1 1 2 ln ( t 2 + 3 t ) d t = 1 2 ln 2 + 1 2 ∫ 1 2 1 ( ln t + ln ( t + 3 ) ) d t = 13 2 ln 2 − 7 4 ln 7 − 1 2 \displaystyle \begin{aligned}\int_{\frac12}^1\ln(2\sqrt{t^2+3t})dt&=\frac12\ln2+\int_{\frac12}^1\frac12\ln(t^2+3t)dt\\&=\frac12\ln2+\frac12\int_{\frac12}^1(\ln t+\ln(t+3))dt=\frac{13}2\ln2-\frac74\ln7-\frac12\end{aligned} ∫ 2 1 1 ln ( 2 t 2 + 3 t ) d t = 2 1 ln 2 + ∫ 2 1 1 2 1 ln ( t 2 + 3 t ) d t = 2 1 ln 2 + 2 1 ∫ 2 1 1 ( ln t + ln ( t + 3 )) d t = 2 13 ln 2 − 4 7 ln 7 − 2 1
+3점 · 20 ln 2 − 5 ln 5 \displaystyle 20\ln2-5\ln5 20 ln 2 − 5 ln 5 ∫ 1 4 ln ( t + 3 t ) d t = [ t ln ( t + 3 t ) ] 1 4 − ∫ 1 4 t 1 + 3 2 t − 1 2 t + 3 t d t = [ t ln ( t + 3 t ) ] 1 4 − ∫ 1 4 t + 3 2 t + 3 d t = 20 ln 2 − 5 ln 5 \displaystyle \begin{aligned}\int_1^4\ln(t+3\sqrt t)dt&=\Biggl[t\ln(t+3\sqrt t)\Biggr]_1^4-\int_1^4t\frac{1+\frac32t^{-\frac12}}{t+3\sqrt t}dt\\&=\Biggl[t\ln(t+3\sqrt t)\Biggr]_1^4-\int_1^4\frac{\sqrt t+\frac32}{\sqrt t+3}dt\\&=20\ln2-5\ln5\end{aligned} ∫ 1 4 ln ( t + 3 t ) d t = [ t ln ( t + 3 t ) ] 1 4 − ∫ 1 4 t t + 3 t 1 + 2 3 t − 2 1 d t = [ t ln ( t + 3 t ) ] 1 4 − ∫ 1 4 t + 3 t + 2 3 d t = 20 ln 2 − 5 ln 5
두 값을 더하면 53 2 ln 2 − 5 ln 5 − 7 4 ln 7 − 1 2 \displaystyle \frac{53}2\ln2-5\ln5-\frac74\ln7-\frac12 2 53 ln 2 − 5 ln 5 − 4 7 ln 7 − 2 1 a = 53 2 \displaystyle a=\frac{53}2 a = 2 53 b = − 5 \displaystyle b=-5 b = − 5 c = − 7 4 \displaystyle c=-\frac74 c = − 4 7 d = − 1 2 \displaystyle d=-\frac12 d = − 2 1
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