integral from 0 to 4 of xe^{x+2}

해법

\int_{0}^{4} x e^{x + 2} d x

3 e^{6} + e^{2}

십진법

1217.67543 \dots

솔루션 단계

\int_{0}^{4} x e^{x + 2} d x

x e^{x + 2} = e^{2} x e^{x}

= \int_{0}^{4} e^{2} x e^{x} d x

정수를 빼라:

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

= e^{2} \cdot \int_{0}^{4} x e^{x} d x

부품별 통합 적용

: [e^{x} x - \int e^{x} d x]_{0}^{4}

= e^{2} [e^{x} x - \int e^{x} d x]_{0}^{4}

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

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

경계 계산:

3 e^{4} + 1

= e^{2} (3 e^{4} + 1)

단순화

= 3 e^{6} + e^{2}