What is the integral from 0 to pi/2 of (y+y^2cos(x)) ?

The integral from 0 to pi/2 of (y+y^2cos(x)) is (piy)/2+y^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 pi/2 of (y+y^2cos(x))

\int_{0}^{\frac{π}{2}} (y + y^{2} cos (x)) d x

\frac{π y}{2} + y^{2}

Solution steps

\int_{0}^{\frac{π}{2}} y + y^{2} cos (x) d x

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}^{\frac{π}{2}} y d x + \int_{0}^{\frac{π}{2}} cos (x) y^{2} d x

\int_{0}^{\frac{π}{2}} y d x = \frac{π y}{2}

\int_{0}^{\frac{π}{2}} cos (x) y^{2} d x = y^{2}

= \frac{π y}{2} + y^{2}

\int_{- 6}^{0} ([\frac{1}{2} x] + ∣ x ∣) d x

\int_{- 1}^{3} ∣ x ∣ - 1 d x

\int_{0}^{5} \frac{1}{25 - t ^{2}} d t

\int_{1}^{3} \frac{1}{x ^{4}} d x

\int_{0}^{1} 5 x + 6 d x

What is the integral from 0 to pi/2 of (y+y^2cos(x)) ?
The integral from 0 to pi/2 of (y+y^2cos(x)) is (piy)/2+y^2