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人気のある微分積分 >

integral from-1 to x of x+1

解

\int_{- 1}^{x} x + 1 d x

解

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

解答ステップ

\int_{- 1}^{x} x + 1 d x

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}^{x} x d x + \int_{- 1}^{x} 1 d x

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

\int_{- 1}^{x} 1 d x = x + 1

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

人気の例

integral from-pi to 0 of (pi+t)cos(nt)

\int_{- π}^{0} (π + t) cos (n t) d t

integral from 500 to 10x+500 of 1/75000

\int_{500}^{10 x + 500} \frac{1}{75000} d y

integral from 1 to 2 of (x+sqrt(x))^2

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

integral from-2 to 1 of x^2-x+2

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

integral from 0 to 2 of x^2-5

\int_{0}^{2} x^{2} - 5 d x