integral from 0 to 1 of 8e^{-2x}

해법

\int_{0}^{1} 8 e^{- 2 x} d x

4 (1 - \frac{1}{e ^{2}})

십진법

3.45865 \dots

솔루션 단계

\int_{0}^{1} 8 e^{- 2 x} d x

정수를 빼라:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{0}^{1} e^{- 2 x} d x

대체 적용

: \int_{0}^{- 2} - \frac{1}{2} e^{u} d u

= 8 \cdot \int_{0}^{- 2} - \frac{1}{2} e^{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 8 (- \int_{- 2}^{0} - \frac{1}{2} e^{u} d u)

정수를 빼라:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 (- (- \frac{1}{2} \cdot \int_{- 2}^{0} e^{u} d u))

공통 적분 사용:

\int e^{u} d u = e^{u}

= 8 (- (- \frac{1}{2} [e^{u}]_{- 2}^{0}))

8 (- (- \frac{1}{2} [e^{u}]_{- 2}^{0}))

간소화하다 :

4 [e^{u}]_{- 2}^{0}

= 4 [e^{u}]_{- 2}^{0}

경계 계산:

1 - \frac{1}{e ^{2}}

= 4 (1 - \frac{1}{e ^{2}})