What is the integral from 0 to t of (y(u)-u) ?

The integral from 0 to t of (y(u)-u) is -(t^2)/2+(yt^2)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t of (y(u)-u)

\int_{0}^{t} (y (u) - u) d u

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

Solution steps

\int_{0}^{t} y u - u d u

Apply the Sum Rule:

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

= - \int_{0}^{t} u d u + \int_{0}^{t} y u d u

\int_{0}^{t} u d u = \frac{t ^{2}}{2}

\int_{0}^{t} y u d u = \frac{y t ^{2}}{2}

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

\int_{0}^{9} π (2 x)^{2} d x

\int_{- 1}^{1} (r + 1)^{2} d r

\int_{0}^{3} \frac{2}{9} x d x

\int_{8}^{11} 4 - 10 e^{- 0.2 x} d x

\int_{- 3}^{4} x - x^{2} + 12 d x

What is the integral from 0 to t of (y(u)-u) ?
The integral from 0 to t of (y(u)-u) is -(t^2)/2+(yt^2)/2