What is the integral from 0 to pi of 5x^2sin(1/2 x) ?

The integral from 0 to pi of 5x^2sin(1/2 x) is 5(8pi-16)

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 of 5x^2sin(1/2 x)

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

5 (8 π - 16)

Decimal Notation

45.66370 \dots

Solution steps

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

Take the constant out:

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

= 5 \cdot \int_{0}^{π} x^{2} sin (\frac{1}{2} x) d x

Simplify

\frac{1}{2} x : \frac{x}{2}

= 5 \cdot \int_{0}^{π} x^{2} sin (\frac{x}{2}) d x

Apply Integration By Parts

: [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

= 5 [- 2 x^{2} cos (\frac{x}{2}) - \int - 4 x cos (\frac{x}{2}) d x]_{0}^{π}

\int - 4 x cos (\frac{x}{2}) d x = - 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))

= 5 [- 2 x^{2} cos (\frac{x}{2}) - (- 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2})))]_{0}^{π}

Simplify

= 5 [- 2 x^{2} cos (\frac{x}{2}) + 16 (\frac{1}{2} x sin (\frac{x}{2}) + cos (\frac{x}{2}))]_{0}^{π}

Compute the boundaries:

8 π - 16

= 5 (8 π - 16)

x \to 1 lim (9 x + 5)

t^{2} y^{^{''}} + t y^{^{'}} + 4 y = sin (2 ln (t))

\int - 5 tan^{5} (4 x) sec^{6} (4 x) d x

\int_{0}^{1} \frac{1}{x ^{0.1}} d x

x \to 0 lim (\frac{tan ( 9 x )}{sin ( 4 x )})

What is the integral from 0 to pi of 5x^2sin(1/2 x) ?
The integral from 0 to pi of 5x^2sin(1/2 x) is 5(8pi-16)