integral from 1 to 3 of (x^3+1/(4x^3))

解答

\int_{1}^{3} (x^{3} + \frac{1}{4 x ^{3}}) d x

\frac{181}{9}

十进制

20.11111 \dots

求解步骤

\int_{1}^{3} x^{3} + \frac{1}{4 x ^{3}} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{3} x^{3} d x + \int_{1}^{3} \frac{1}{4 x ^{3}} d x

\int_{1}^{3} x^{3} d x = 20

\int_{1}^{3} \frac{1}{4 x ^{3}} d x = \frac{1}{9}

= 20 + \frac{1}{9}

化简

20 + \frac{1}{9} : \frac{181}{9}

= \frac{181}{9}

n or ma l y = 2 x - 1 + 4 x^{2}, (1, 5)

\int k e^{- x} d x

\frac{\partial}{\partial x} (y (e^{x} + x e^{x}))

d er i v a t i v e g (x) = \frac{1}{4 x ^{2} + 2 x}

l a pl a ce t r an s f or m 2 t + 3 t \cdot cos (2 t)