ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 0 to 4 of (xcos(x^2))/2

解

\int_{0}^{4} \frac{x cos ( x ^{2} )}{2} d x

解

\frac{1}{4} sin (16)

+1

十進法表記

- 0.07197 \dots

解答ステップ

\int_{0}^{4} \frac{x cos ( x ^{2} )}{2} d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{4} x cos (x^{2}) d x

u置換積分法を適用する

= \frac{1}{2} \cdot \int_{0}^{16} \frac{cos ( u )}{2} d u

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{16} cos (u) d u

共通積分を使用する:

\int cos (u) d u = sin (u)

= \frac{1}{2} \cdot \frac{1}{2} [sin (u)]_{0}^{16}

簡素化

\frac{1}{2} \cdot \frac{1}{2} [sin (u)]_{0}^{16} : \frac{1}{4} [sin (u)]_{0}^{16}

= \frac{1}{4} [sin (u)]_{0}^{16}

限度を計算する:

sin (16)

= \frac{1}{4} sin (16)

グラフ

人気の例

integral from 1 to 4 of 1/(p^2+p-6)

\int_{1}^{4} \frac{1}{p ^{2} + p - 6} d p

integral from 1 to x of 8/(3t^3)

\int_{1}^{x} \frac{8}{3 t ^{3}} d t

integral from 0 to 1 of 4x^3-2x+6

\int_{0}^{1} 4 x^{3} - 2 x + 6 d x

integral from 1 to 6 of sqrt(5x-5)

\int_{1}^{6} 5 x - 5 d x

integral from-2 to 2 of 6sqrt(4-x^2)

\int_{- 2}^{2} 6 4 - x^{2} d x