laplacetransform 9t

解答

拉普拉斯变换

9 t

\frac{9}{s ^{2}}

求解步骤

L {9 t}

利用拉普拉斯变换的常数乘法特性:

对于函数

f (t)

和常数

a : L {a \cdot f (t)} = a \cdot L {f (t)}

= 9 L {t}

L {t} : \frac{1}{s ^{2}}

= 9 \cdot \frac{1}{s ^{2}}

整理

9 \frac{1}{s ^{2}} : \frac{9}{s ^{2}}

= \frac{9}{s ^{2}}

a re a f (x) = (x - 1)^{2} + 2, x = - 1, x = 2

\int \frac{5 x ^{2} - 12 x - 12}{x ^{3} - 4 x} d x

\int \frac{2 x ^{5} - x ^{3} - 1}{x ^{3} - 4 x} d x

x \to 1 + lim (\frac{2 x}{x - 1})

\int_{0}^{2} (2 x^{2} + 1) d x