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人気のある微分積分 >

integral from 1 to t of-t+2

解

\int_{1}^{t} - t + 2 d t

解

- \frac{t ^{2} - 1}{2} + 2 t - 2

解答ステップ

\int_{1}^{t} - t + 2 d t

総和規則を適用する:

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

= - \int_{1}^{t} t d t + \int_{1}^{t} 2 d t

\int_{1}^{t} t d t = \frac{t ^{2} - 1}{2}

\int_{1}^{t} 2 d t = 2 t - 2

= - \frac{t ^{2} - 1}{2} + 2 t - 2

人気の例

laplacetransform 12te^{10t}

l a pl a ce t r an s f or m 12 t e^{10 t}

\frac{d y}{d x} - 2 = - y

d/(dy)(ycos(y))

\frac{d}{d y} (y cos (y))

derivative f(x)=(x^2)/(x-6)

d er i v a t i v e f (x) = \frac{x ^{2}}{x - 6}

x*(dy)/(dx)+y=-6xy^2,y(1)=-7

x \cdot \frac{d y}{d x} + y = - 6 x y^{2}, y (1) = - 7