정의 우함수와 기함수
모든 실수 x x x f ( − x ) = f ( x ) f(-x)=f(x) f ( − x ) = f ( x ) 우함수 라 한다.y y y 대칭 이다.f ( − x ) = − f ( x ) f(-x)=-f(x) f ( − x ) = − f ( x ) 기함수 라 한다. 이전 개념 성질 ∫ − a a \displaystyle\int_{-a}^{a} ∫ − a a 연속함수 f ( x ) f(x) f ( x ) f ( − x ) = f ( x ) f(-x)=f(x) f ( − x ) = f ( x ) ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x \int_{-a}^{a} f(x)\,dx=2\int_{0}^{a} f(x)\,dx ∫ − a a f ( x ) d x = 2 ∫ 0 a f ( x ) d x f ( − x ) = − f ( x ) f(-x)=-f(x) f ( − x ) = − f ( x ) ∫ − a a f ( x ) d x = 0 \int_{-a}^{a} f(x)\,dx=0 ∫ − a a f ( x ) d x = 0 y y y x 2 x^2 x 2 x 4 x^4 x 4 x x x x 3 x^3 x 3
학생 아래끝과 위끝이 아무 수여도 이렇게 계산해요?
쌤 아니요. − a -a − a a a a ∫ − 1 2 \displaystyle\int_{-1}^{2} ∫ − 1 2 다항함수의 ∫ − a a \displaystyle\int_{-a}^{a} ∫ − a a 1 2 3 ∫ − 1 1 ( x 3 + 3 x 2 + x + 1 ) d x \displaystyle\int_{-1}^{1} (x^3+3x^2+x+1)\,dx ∫ − 1 1 ( x 3 + 3 x 2 + x + 1 ) d x = 2 ∫ 0 1 ( 3 x 2 + 1 ) d x =2\int_{0}^{1} (3x^2+1)\,dx = 2 ∫ 0 1 ( 3 x 2 + 1 ) d x ↳ ◀ x 3 x^3 x 3 x x x = 2 [ x 3 + x ] 0 1 = 2 × 2 = 4 =2\Big[x^3+x\Big]_{0}^{1}=2\times2=4 = 2 [ x 3 + x ] 0 1 = 2 × 2 = 4 바로 확인 방금 배운 것, 10초 문제
∫ − 2 2 ( x 3 + x ) d x \displaystyle\int_{-2}^{2}(x^3+x)\,dx ∫ − 2 2 ( x 3 + x ) d x 홀수 차수 항까지 두 배 한 실수
✗ 틀린 풀이
∫ − 1 1 ( x 3 + 3 x 2 + x + 1 ) d x = 2 ∫ 0 1 ( x 3 + 3 x 2 + x + 1 ) d x = 11 2 \displaystyle\int_{-1}^{1} (x^3+3x^2+x+1)\,dx=2\int_{0}^{1} (x^3+3x^2+x+1)\,dx=\dfrac{11}{2} ∫ − 1 1 ( x 3 + 3 x 2 + x + 1 ) d x = 2 ∫ 0 1 ( x 3 + 3 x 2 + x + 1 ) d x = 2 11 왜 틀렸나 ∫ − a a \displaystyle\int_{-a}^{a} ∫ − a a 0 0 0 ↓ ✓ 바른 풀이
∫ − 1 1 ( x 3 + 3 x 2 + x + 1 ) d x = 2 ∫ 0 1 ( 3 x 2 + 1 ) d x = 4 \displaystyle\int_{-1}^{1} (x^3+3x^2+x+1)\,dx=2\int_{0}^{1} (3x^2+1)\,dx=4 ∫ − 1 1 ( x 3 + 3 x 2 + x + 1 ) d x = 2 ∫ 0 1 ( 3 x 2 + 1 ) d x = 4 계산: 2 ∫ 0 1 ( 3 x 2 + 1 ) d x = 2 ( 1 + 1 ) = 4 2\displaystyle\int_{0}^{1}(3x^2+1)\,dx=2(1+1)=4 2 ∫ 0 1 ( 3 x 2 + 1 ) d x = 2 ( 1 + 1 ) = 4 y = x 3 + x y=x^3+x y = x 3 + x ∫ − 1 1 ( x 3 + x ) d x = 0 \displaystyle\int_{-1}^{1}(x^3+x)\,dx=0 ∫ − 1 1 ( x 3 + x ) d x = 0 + 펼치기 접기 예제 왜 그런가 ∫ − a a x n d x \displaystyle\int_{-a}^{a} x^n\,dx ∫ − a a x n d x ∫ − a a x n d x = [ x n + 1 n + 1 ] − a a \int_{-a}^{a} x^n\,dx=\Big[\dfrac{x^{n+1}}{n+1}\Big]_{-a}^{a} ∫ − a a x n d x = [ n + 1 x n + 1 ] − a a ↳ ◀ 부정적분에 위끝·아래끝 넣기
a n + 1 n + 1 − ( − a ) n + 1 n + 1 = 2 a n + 1 n + 1 \dfrac{a^{n+1}}{n+1}-\dfrac{(-a)^{n+1}}{n+1}=\dfrac{2a^{n+1}}{n+1} n + 1 a n + 1 − n + 1 ( − a ) n + 1 = n + 1 2 a n + 1 ↳ ◀ n n n ( − a ) n + 1 = − a n + 1 (-a)^{n+1}=-a^{n+1} ( − a ) n + 1 = − a n + 1 = 2 ∫ 0 a x n d x =2\int_{0}^{a} x^n\,dx = 2 ∫ 0 a x n d x a n + 1 n + 1 − ( − a ) n + 1 n + 1 = 0 \dfrac{a^{n+1}}{n+1}-\dfrac{(-a)^{n+1}}{n+1}=0 n + 1 a n + 1 − n + 1 ( − a ) n + 1 = 0 ↳ ◀ n n n ( − a ) n + 1 = a n + 1 (-a)^{n+1}=a^{n+1} ( − a ) n + 1 = a n + 1 ∫ 1 x t 2 d t = [ 1 3 t 3 ] 1 x = 1 3 x 3 − 1 3 \displaystyle\int_{1}^{x} t^2\,dt=\Big[\dfrac{1}{3}t^3\Big]_{1}^{x}=\dfrac{1}{3}x^3-\dfrac{1}{3} ∫ 1 x t 2 d t = [ 3 1 t 3 ] 1 x = 3 1 x 3 − 3 1 x x x 정적분 은 x x x x x x t t t x x x
이전 개념 성질 정적분으로 정의된 함수의 미분
함수 f ( t ) f(t) f ( t ) a a a d d x ∫ a x f ( t ) d t = f ( x ) \dfrac{d}{dx}\int_{a}^{x} f(t)\,dt=f(x) d x d ∫ a x f ( t ) d t = f ( x ) 미적분의 기본정리 로 계산해도 같다. f ( t ) f(t) f ( t ) F ( t ) F(t) F ( t ) ∫ a x f ( t ) d t = F ( x ) − F ( a ) \displaystyle\int_{a}^{x} f(t)\,dt=F(x)-F(a) ∫ a x f ( t ) d t = F ( x ) − F ( a ) F ( a ) F(a) F ( a ) F ′ ( x ) = f ( x ) F'(x)=f(x) F ′ ( x ) = f ( x ) 이전 개념 d d x ∫ 1 x ( t 2 − 3 t ) d t \dfrac{d}{dx}\displaystyle\int_{1}^{x} (t^2-3t)\,dt d x d ∫ 1 x ( t 2 − 3 t ) d t d d x ∫ 1 x ( t 2 − 3 t ) d t = x 2 − 3 x \dfrac{d}{dx}\int_{1}^{x} (t^2-3t)\,dt=x^2-3x d x d ∫ 1 x ( t 2 − 3 t ) d t = x 2 − 3 x ∫ 1 x ( t 2 − 3 t ) d t = 1 3 x 3 − 3 2 x 2 + 7 6 \int_{1}^{x} (t^2-3t)\,dt=\dfrac{1}{3}x^3-\dfrac{3}{2}x^2+\dfrac{7}{6} ∫ 1 x ( t 2 − 3 t ) d t = 3 1 x 3 − 2 3 x 2 + 6 7 ↳ ◀ 확인: 직접 계산
( 1 3 x 3 − 3 2 x 2 + 7 6 ) ′ = x 2 − 3 x \left(\dfrac{1}{3}x^3-\dfrac{3}{2}x^2+\dfrac{7}{6}\right)'=x^2-3x ( 3 1 x 3 − 2 3 x 2 + 6 7 ) ′ = x 2 − 3 x ↳ ◀ 미분하면 같아요
쌤 ∫ 1 x f ( t ) d t = F ( x ) − F ( 1 ) \displaystyle\int_{1}^{x} f(t)\,dt=F(x)-F(1) ∫ 1 x f ( t ) d t = F ( x ) − F ( 1 ) F ( 1 ) F(1) F ( 1 ) 0 0 0 f ( x ) f(x) f ( x ) 위끝과 아래끝이 같아지므로 ∫ a a f ( t ) d t = 0 \int_{a}^{a} f(t)\,dt=0 ∫ a a f ( t ) d t = 0 바로 확인 방금 배운 것, 10초 문제
d d x ∫ 2 x ( t 3 − t ) d t \dfrac{d}{dx}\displaystyle\int_{2}^{x} (t^3-t)\,dt d x d ∫ 2 x ( t 3 − t ) d t 3 x 2 − 1 3x^2-1 3 x 2 − 1 x 3 − x − 6 x^3-x-6 x 3 − x − 6 x 3 − x x^3-x x 3 − x 아래끝의 값을 빼는 실수
✗ 틀린 풀이
d d x ∫ 1 x ( t 2 − 3 t ) d t = ( x 2 − 3 x ) − ( 1 − 3 ) = x 2 − 3 x + 2 \dfrac{d}{dx}\displaystyle\int_{1}^{x} (t^2-3t)\,dt=(x^2-3x)-(1-3)=x^2-3x+2 d x d ∫ 1 x ( t 2 − 3 t ) d t = ( x 2 − 3 x ) − ( 1 − 3 ) = x 2 − 3 x + 2 왜 틀렸나 위끝이 x x x x x x x x x ↓ ✓ 바른 풀이
d d x ∫ 1 x ( t 2 − 3 t ) d t = x 2 − 3 x \dfrac{d}{dx}\displaystyle\int_{1}^{x} (t^2-3t)\,dt=x^2-3x d x d ∫ 1 x ( t 2 − 3 t ) d t = x 2 − 3 x 아래끝 1 1 1 F ( 1 ) F(1) F ( 1 ) 0 0 0 ∫ 1 x ( t 2 − 3 t ) d t = x 3 3 − 3 x 2 2 + 7 6 \displaystyle\int_{1}^{x}(t^2-3t)\,dt=\dfrac{x^3}{3}-\dfrac{3x^2}{2}+\dfrac{7}{6} ∫ 1 x ( t 2 − 3 t ) d t = 3 x 3 − 2 3 x 2 + 6 7 x 2 − 3 x x^2-3x x 2 − 3 x d d x ∫ a x f ( t ) d t = f ( x ) \displaystyle \dfrac{d}{dx}\int_{a}^{x} f(t)\,dt=f(x) d x d ∫ a x f ( t ) d t = f ( x ) 위끝·아래끝이 상수
∫ 0 1 f ( t ) d t = k \displaystyle \int_{0}^{1} f(t)\,dt=k ∫ 0 1 f ( t ) d t = k 하나의 수(상수)이므로 미분하면 0 0 0 예 위끝·아래끝이 상수인 정적분이 들어 있는 경우
f ( x ) = x + 2 ∫ 0 1 f ( t ) d t f(x)=x+2\displaystyle\int_{0}^{1} f(t)\,dt f ( x ) = x + 2 ∫ 0 1 f ( t ) d t ∫ 0 1 f ( t ) d t \displaystyle\int_{0}^{1} f(t)\,dt ∫ 0 1 f ( t ) d t 상수 이므로 k k k f ( x ) = x + 2 k f(x)=x+2k f ( x ) = x + 2 k k = ∫ 0 1 ( t + 2 k ) d t = 1 2 + 2 k k=\displaystyle\int_{0}^{1} (t+2k)\,dt=\dfrac{1}{2}+2k k = ∫ 0 1 ( t + 2 k ) d t = 2 1 + 2 k k = − 1 2 k=-\dfrac{1}{2} k = − 2 1 f ( x ) = x − 1 f(x)=x-1 f ( x ) = x − 1 ∫ a x f ( t ) d t = g ( x ) \displaystyle\int_{a}^{x} f(t)\,dt=g(x) ∫ a x f ( t ) d t = g ( x ) x = a x=a x = a 0 0 0 g ( a ) = 0 g(a)=0 g ( a ) = 0 f ( x ) = g ′ ( x ) f(x)=g'(x) f ( x ) = g ′ ( x )
두 가지 조건을 쓴다
1 왼쪽은 ∫ a a f ( t ) d t = 0 \displaystyle\int_{a}^{a} f(t)\,dt=0 ∫ a a f ( t ) d t = 0 g ( a ) = 0 g(a)=0 g ( a ) = 0 a a a 2 왼쪽은 f ( x ) f(x) f ( x ) f ( x ) = g ′ ( x ) f(x)=g'(x) f ( x ) = g ′ ( x ) 3 ∫ a x f ( t ) d t = x 2 − 3 x + 2 \displaystyle\int_{a}^{x} f(t)\,dt=x^2-3x+2 ∫ a x f ( t ) d t = x 2 − 3 x + 2 0 = a 2 − 3 a + 2 0=a^2-3a+2 0 = a 2 − 3 a + 2 ( a − 1 ) ( a − 2 ) = 0 , a = 1 또는 a = 2 (a-1)(a-2)=0,\ a=1\ \text{또는}\ a=2 ( a − 1 ) ( a − 2 ) = 0 , a = 1 또는 a = 2 f ( x ) = 2 x − 3 f(x)=2x-3 f ( x ) = 2 x − 3 쌤 네. a = 1 a=1 a = 1 a = 2 a=2 a = 2 ∫ a x ( 2 t − 3 ) d t = x 2 − 3 x + 2 \displaystyle\int_{a}^{x} (2t-3)\,dt=x^2-3x+2 ∫ a x ( 2 t − 3 ) d t = x 2 − 3 x + 2 바로 확인 방금 배운 것, 10초 문제
∫ 1 x f ( t ) d t = x 2 + k x \displaystyle\int_{1}^{x} f(t)\,dt=x^2+kx ∫ 1 x f ( t ) d t = x 2 + k x k k k ✗ 틀린 풀이
∫ a x f ( t ) d t = x 2 − 3 x + 2 \displaystyle\int_{a}^{x} f(t)\,dt=x^2-3x+2 ∫ a x f ( t ) d t = x 2 − 3 x + 2 f ( x ) = 2 x − 3 f(x)=2x-3 f ( x ) = 2 x − 3 a a a 왜 틀렸나 위끝과 아래끝이 같으면 정적분은 0 0 0 x = a x=a x = a a a a ↓ ✓ 바른 풀이
x = a x=a x = a a 2 − 3 a + 2 = 0 a^2-3a+2=0 a 2 − 3 a + 2 = 0 a = 1 a=1 a = 1 a = 2 a=2 a = 2 확인: a = 1 a=1 a = 1 ∫ 1 x ( 2 t − 3 ) d t = x 2 − 3 x + 2 \displaystyle\int_{1}^{x}(2t-3)\,dt=x^2-3x+2 ∫ 1 x ( 2 t − 3 ) d t = x 2 − 3 x + 2 a = 2 a=2 a = 2 + 펼치기 접기 예제 응용 ∫ 1 x f ( t ) d t = x 3 − 2 x 2 + k \displaystyle\int_{1}^{x} f(t)\,dt=x^3-2x^2+k ∫ 1 x f ( t ) d t = x 3 − 2 x 2 + k 0 = 1 − 2 + k , k = 1 0=1-2+k,\ k=1 0 = 1 − 2 + k , k = 1 f ( x ) = 3 x 2 − 4 x f(x)=3x^2-4x f ( x ) = 3 x 2 − 4 x F ( x ) = ∫ a x f ( t ) d t F(x)=\displaystyle\int_{a}^{x} f(t)\,dt F ( x ) = ∫ a x f ( t ) d t F ′ ( x ) = f ( x ) F'(x)=f(x) F ′ ( x ) = f ( x ) F ( x ) F(x) F ( x ) 극값 은 피적분함수 f ( x ) f(x) f ( x ) F ( x ) F(x) F ( x ) x x x
이전 개념 F ( x ) = ∫ a x f ( t ) d t F(x)=\displaystyle\int_{a}^{x} f(t)\,dt F ( x ) = ∫ a x f ( t ) d t 1 2 부호가 + + + − - − − - − + + + 3 F ( x ) = ∫ 0 x ( t 2 − 4 t + 3 ) d t F(x)=\displaystyle\int_{0}^{x} (t^2-4t+3)\,dt F ( x ) = ∫ 0 x ( t 2 − 4 t + 3 ) d t F ′ ( x ) F'(x) F ′ ( x ) x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 ) x^2-4x+3=(x-1)(x-3) x 2 − 4 x + 3 = ( x − 1 ) ( x − 3 ) F ( x ) F(x) F ( x ) 1 3 x 3 − 2 x 2 + 3 x \dfrac{1}{3}x^3-2x^2+3x 3 1 x 3 − 2 x 2 + 3 x 극댓값
F ( 1 ) = 1 3 − 2 + 3 = 4 3 F(1)=\dfrac{1}{3}-2+3=\dfrac{4}{3} F ( 1 ) = 3 1 − 2 + 3 = 3 4 , 극솟값
F ( 3 ) = 9 − 18 + 9 = 0 F(3)=9-18+9=0 F ( 3 ) = 9 − 18 + 9 = 0 학생 극값을 구할 때도 F ( x ) F(x) F ( x ) 쌤 극값을 갖는 x x x f ( x ) f(x) f ( x ) x x x F ( 1 ) F(1) F ( 1 ) F ( 3 ) F(3) F ( 3 ) 바로 확인 방금 배운 것, 10초 문제
F ( x ) = ∫ 0 x ( t − 2 ) d t F(x)=\displaystyle\int_{0}^{x} (t-2)\,dt F ( x ) = ∫ 0 x ( t − 2 ) d t 극값 대신 피적분함수의 값을 답한 실수
✗ 틀린 풀이
x = 1 x=1 x = 1 f ( 1 ) = 1 − 4 + 3 = 0 f(1)=1-4+3=0 f ( 1 ) = 1 − 4 + 3 = 0 왜 틀렸나 극댓값은 F ′ ( x ) = f ( x ) = 0 F'(x)=f(x)=0 F ′ ( x ) = f ( x ) = 0 x = 1 x=1 x = 1 F ( x ) F(x) F ( x ) ↓ ✓ 바른 풀이
극댓값은 F ( 1 ) = ∫ 0 1 ( t 2 − 4 t + 3 ) d t = 4 3 F(1)=\displaystyle\int_{0}^{1} (t^2-4t+3)\,dt=\dfrac{4}{3} F ( 1 ) = ∫ 0 1 ( t 2 − 4 t + 3 ) d t = 3 4 계산: F ( 1 ) = [ t 3 3 − 2 t 2 + 3 t ] 0 1 = 1 3 − 2 + 3 = 4 3 F(1)=\Big[\dfrac{t^3}{3}-2t^2+3t\Big]_{0}^{1}=\dfrac{1}{3}-2+3=\dfrac{4}{3} F ( 1 ) = [ 3 t 3 − 2 t 2 + 3 t ] 0 1 = 3 1 − 2 + 3 = 3 4 성질 정적분으로 나타낸 극한
f ( x ) f(x) f ( x ) F ′ ( x ) = f ( x ) F'(x)=f(x) F ′ ( x ) = f ( x ) lim x → a 1 x − a ∫ a x f ( t ) d t = f ( a ) \lim_{x\to a}\dfrac{1}{x-a}\int_{a}^{x} f(t)\,dt=f(a) x → a lim x − a 1 ∫ a x f ( t ) d t = f ( a ) lim h → 0 1 h ∫ a a + h f ( t ) d t = f ( a ) \lim_{h\to 0}\dfrac{1}{h}\int_{a}^{a+h} f(t)\,dt=f(a) h → 0 lim h 1 ∫ a a + h f ( t ) d t = f ( a ) lim x → a F ( x ) − F ( a ) x − a \displaystyle\lim_{x\to a}\dfrac{F(x)-F(a)}{x-a} x → a lim x − a F ( x ) − F ( a ) 미분계수 F ′ ( a ) F'(a) F ′ ( a ) F ′ ( a ) = f ( a ) F'(a)=f(a) F ′ ( a ) = f ( a ) lim h → 0 F ( a + h ) − F ( a ) h = F ′ ( a ) \displaystyle\lim_{h\to 0}\dfrac{F(a+h)-F(a)}{h}=F'(a) h → 0 lim h F ( a + h ) − F ( a ) = F ′ ( a ) 이전 개념 예 lim x → 1 1 x − 1 ∫ 1 x ( t 3 + 2 t ) d t \displaystyle\lim_{x\to 1}\dfrac{1}{x-1}\int_{1}^{x} (t^3+2t)\,dt x → 1 lim x − 1 1 ∫ 1 x ( t 3 + 2 t ) d t f ( t ) = t 3 + 2 t f(t)=t^3+2t f ( t ) = t 3 + 2 t f ( 1 ) f(1) f ( 1 ) f ( 1 ) = 1 + 2 = 3 f(1)=1+2=3 f ( 1 ) = 1 + 2 = 3 학생 lim h → 0 1 h ∫ a a + 2 h f ( t ) d t \displaystyle\lim_{h\to0}\dfrac{1}{h}\int_{a}^{a+2h} f(t)\,dt h → 0 lim h 1 ∫ a a + 2 h f ( t ) d t a + 2 h a+2h a + 2 h 쌤 분모를 2 h 2h 2 h 1 h = 2 × 1 2 h \dfrac{1}{h}=2\times\dfrac{1}{2h} h 1 = 2 × 2 h 1 2 f ( a ) 2f(a) 2 f ( a ) 바로 확인 방금 배운 것, 10초 문제
lim x → 2 1 x − 2 ∫ 2 x ( t 2 + 1 ) d t \displaystyle\lim_{x\to2}\dfrac{1}{x-2}\int_{2}^{x} (t^2+1)\,dt x → 2 lim x − 2 1 ∫ 2 x ( t 2 + 1 ) d t ✗ 틀린 풀이
h → 0 h\to 0 h → 0 0 0 0 왜 틀렸나 분자와 분모가 함께 0 0 0 0 0 \dfrac{0}{0} 0 0 ↓ ✓ 바른 풀이
lim h → 0 1 h ∫ 1 1 + h ( t 3 + 2 t ) d t = f ( 1 ) = 3 \displaystyle\lim_{h\to 0}\dfrac{1}{h}\int_{1}^{1+h} (t^3+2t)\,dt=f(1)=3 h → 0 lim h 1 ∫ 1 1 + h ( t 3 + 2 t ) d t = f ( 1 ) = 3 F ( x ) = ∫ 1 x f ( t ) d t F(x)=\displaystyle\int_{1}^{x}f(t)\,dt F ( x ) = ∫ 1 x f ( t ) d t F ( 1 + h ) − F ( 1 ) h \dfrac{F(1+h)-F(1)}{h} h F ( 1 + h ) − F ( 1 ) F ′ ( 1 ) = f ( 1 ) = 1 + 2 = 3 F'(1)=f(1)=1+2=3 F ′ ( 1 ) = f ( 1 ) = 1 + 2 = 3 + 펼치기 접기 예제 응용 lim h → 0 1 h ∫ 1 1 + 2 h ( t 3 + 2 t ) d t \displaystyle\lim_{h\to 0}\dfrac{1}{h}\int_{1}^{1+2h} (t^3+2t)\,dt h → 0 lim h 1 ∫ 1 1 + 2 h ( t 3 + 2 t ) d t lim h → 0 F ( 1 + 2 h ) − F ( 1 ) h \lim_{h\to 0}\dfrac{F(1+2h)-F(1)}{h} lim h → 0 h F ( 1 + 2 h ) − F ( 1 ) ↳ ◀ F ′ ( t ) = t 3 + 2 t F'(t)=t^3+2t F ′ ( t ) = t 3 + 2 t = lim h → 0 F ( 1 + 2 h ) − F ( 1 ) 2 h × 2 =\lim_{h\to 0}\dfrac{F(1+2h)-F(1)}{2h}\times2 = lim h → 0 2 h F ( 1 + 2 h ) − F ( 1 ) × 2 = 2 F ′ ( 1 ) = 2 f ( 1 ) = 6 =2F'(1)=2f(1)=6 = 2 F ′ ( 1 ) = 2 f ( 1 ) = 6