+5점 · 곡선 y = f ( x ) \displaystyle y=f(x) y = f ( x ) ( t , f ( t ) ) \displaystyle (t,f(t)) ( t , f ( t )) y − ( t − 2 ) 2 e t = t ( t − 2 ) e t ( x − t ) \displaystyle y-(t-2)^2e^t=t(t-2)e^t(x-t) y − ( t − 2 ) 2 e t = t ( t − 2 ) e t ( x − t ) +5점 · 두 접선 y = 0 \displaystyle y=0 y = 0 y = − e x + 2 e \displaystyle y=-ex+2e y = − e x + 2 e (1-1) 곡선 y = f ( x ) \displaystyle y=f(x) y = f ( x ) ( t , f ( t ) ) \displaystyle (t,f(t)) ( t , f ( t )) y − ( t − 2 ) 2 e t = t ( t − 2 ) e t ( x − t ) \displaystyle y-(t-2)^2e^t=t(t-2)e^t(x-t) y − ( t − 2 ) 2 e t = t ( t − 2 ) e t ( x − t ) 이다. 이 직선이 ( 2 , 0 ) \displaystyle (2,0) ( 2 , 0 ) t = 1 \displaystyle t=1 t = 1 t = 2 \displaystyle t=2 t = 2 따라서 구하고자 하는 직선은 y = 0 , y = − e x + 2 e \displaystyle y=0,\qquad y=-ex+2e y = 0 , y = − e x + 2 e 이다.
+5점 · ∫ ( x − 2 ) 2 e x d x = ( x 2 − 6 x + 10 ) e x + C \displaystyle \int(x-2)^2e^xdx=(x^2-6x+10)e^x+C ∫ ( x − 2 ) 2 e x d x = ( x 2 − 6 x + 10 ) e x + C +5점 · S = 16 − 2 e 2 \displaystyle S=16-2e^2 S = 16 − 2 e 2 (1-2) f ′ ( x ) = x ( x − 2 ) e x \displaystyle f^{\prime}(x)=x(x-2)e^x f ′ ( x ) = x ( x − 2 ) e x a = 0 \displaystyle a=0 a = 0 b = 2 \displaystyle b=2 b = 2 ∫ x e x d x = ( x − 1 ) e x + C \displaystyle \int xe^xdx=(x-1)e^x+C ∫ x e x d x = ( x − 1 ) e x + C ∫ ( x − 2 ) 2 e x d x = ( x 2 − 6 x + 10 ) e x + C \displaystyle \int(x-2)^2e^xdx=(x^2-6x+10)e^x+C ∫ ( x − 2 ) 2 e x d x = ( x 2 − 6 x + 10 ) e x + C 이다. 그러므로 S = ∫ 0 2 [ 6 − 3 x − ( x − 2 ) 2 e x ] d x = [ 6 x − 3 2 x 2 − ( x 2 − 6 x + 10 ) e x ] 0 2 = 16 − 2 e 2 \displaystyle S=\int_0^2[6-3x-(x-2)^2e^x]dx=\Biggl[6x-\frac{3}{2}x^2-(x^2-6x+10)e^x\Biggr]_0^2=16-2e^2 S = ∫ 0 2 [ 6 − 3 x − ( x − 2 ) 2 e x ] d x = [ 6 x − 2 3 x 2 − ( x 2 − 6 x + 10 ) e x ] 0 2 = 16 − 2 e 2 이다.
(1-3) S \displaystyle S S y = g ( x ) \displaystyle y=g(x) y = g ( x ) y = f ( x ) \displaystyle y=f(x) y = f ( x ) g ( x ) = f ( t ) + f ′ ( t ) ( x − t ) \displaystyle g(x)=f(t)+f^{\prime}(t)(x-t) g ( x ) = f ( t ) + f ′ ( t ) ( x − t ) y = f ( x ) \displaystyle y=f(x) y = f ( x ) − 1 ≤ t ≤ 1 \displaystyle -1\le t\le1 − 1 ≤ t ≤ 1
+5점 · g ( x ) = f ( 1 2 ) + f ′ ( 1 2 ) ( x − 1 2 ) \displaystyle g(x)=f\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)\left(x-\frac{1}{2}\right) g ( x ) = f ( 2 1 ) + f ′ ( 2 1 ) ( x − 2 1 ) S \displaystyle S S +5점 · S \displaystyle S S 27 4 e − 2 e 2 + 17 e \displaystyle \frac{27}{4}\sqrt{e}-2e^2+\frac{17}{e} 4 27 e − 2 e 2 + e 17 한편 g ( x ) = f ( t ) + f ′ ( t ) ( x − t ) \displaystyle g(x)=f(t)+f^{\prime}(t)(x-t) g ( x ) = f ( t ) + f ′ ( t ) ( x − t ) S \displaystyle S S h ( t ) \displaystyle h(t) h ( t ) h ( t ) = ∫ − 1 2 [ f ( t ) + f ′ ( t ) ( x − t ) − f ( x ) ] d x = 3 f ( t ) + ( − 3 t + 3 2 ) f ′ ( t ) − ∫ − 1 2 f ( x ) d x \displaystyle h(t)=\int_{-1}^2[f(t)+f^{\prime}(t)(x-t)-f(x)]dx=3f(t)+\left(-3t+\frac{3}{2}\right)f^{\prime}(t)-\int_{-1}^2 f(x)dx h ( t ) = ∫ − 1 2 [ f ( t ) + f ′ ( t ) ( x − t ) − f ( x )] d x = 3 f ( t ) + ( − 3 t + 2 3 ) f ′ ( t ) − ∫ − 1 2 f ( x ) d x 이고 d h d t = ( − 3 t + 3 2 ) f ′ ′ ( t ) = − 3 ( t − 1 2 ) ( t 2 − 2 ) e t \displaystyle \frac{dh}{dt}=\left(-3t+\frac{3}{2}\right)f^{\prime\prime}(t)=-3\left(t-\frac{1}{2}\right)(t^2-2)e^t d t d h = ( − 3 t + 2 3 ) f ′′ ( t ) = − 3 ( t − 2 1 ) ( t 2 − 2 ) e t 이다. 따라서 S \displaystyle S S t = 1 2 \displaystyle t=\frac{1}{2} t = 2 1 3 f ( 1 2 ) − ∫ − 1 2 f ( x ) d x = 27 4 e − [ ( x 2 − 6 x + 10 ) e x ] − 1 2 = 27 4 e − 2 e 2 + 17 e \displaystyle 3f\left(\frac{1}{2}\right)-\int_{-1}^2 f(x)dx=\frac{27}{4}\sqrt{e}-\Biggl[(x^2-6x+10)e^x\Biggr]_{-1}^2=\frac{27}{4}\sqrt{e}-2e^2+\frac{17}{e} 3 f ( 2 1 ) − ∫ − 1 2 f ( x ) d x = 4 27 e − [ ( x 2 − 6 x + 10 ) e x ] − 1 2 = 4 27 e − 2 e 2 + e 17 을 갖는다.
수리논술, 배움에서 논증의 완성까지.