What is the integral from 0 to pi/3 of x^2sin(3x) ?

The integral from 0 to pi/3 of x^2sin(3x) is (pi^2-4)/(27)

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/3 of x^2sin(3x)

\int_{0}^{\frac{π}{3}} x^{2} sin (3 x) d x

\frac{π ^{2} - 4}{27}

Decimal

0.21739 \dots

Solution steps

\int_{0}^{\frac{π}{3}} x^{2} sin (3 x) d x

Apply Integration By Parts

: [\frac{1}{3} (- x^{2} cos (3 x) - 3 \cdot \int - \frac{2}{3} x cos (3 x) d x)]_{0}^{\frac{π}{3}}

= [\frac{1}{3} (- x^{2} cos (3 x) - 3 \cdot \int - \frac{2}{3} x cos (3 x) d x)]_{0}^{\frac{π}{3}}

\int - \frac{2}{3} x cos (3 x) d x = - \frac{2}{27} (3 x sin (3 x) + cos (3 x))

= [\frac{1}{3} (- x^{2} cos (3 x) - 3 (- \frac{2}{27} (3 x sin (3 x) + cos (3 x))))]_{0}^{\frac{π}{3}}

Simplify

[\frac{1}{3} (- x^{2} cos (3 x) - 3 (- \frac{2}{27} (3 x sin (3 x) + cos (3 x))))]_{0}^{\frac{π}{3}} : [\frac{1}{3} (- x^{2} cos (3 x) + \frac{2}{9} (3 x sin (3 x) + cos (3 x)))]_{0}^{\frac{π}{3}}

= [\frac{1}{3} (- x^{2} cos (3 x) + \frac{2}{9} (3 x sin (3 x) + cos (3 x)))]_{0}^{\frac{π}{3}}

Compute the boundaries:

\frac{π ^{2} - 2}{27} - \frac{2}{27}

= \frac{π ^{2} - 2}{27} - \frac{2}{27}

Simplify

= \frac{π ^{2} - 4}{27}

\int_{1}^{l n (8)} 6 x + e^{x} d x

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

\int_{0}^{1} \frac{1}{x - x ^{2}} d x

\int_{- π}^{0} x cos (n x) d x

\int_{0}^{\frac{π}{4}} cos^{7} (2 x) d x

What is the integral from 0 to pi/3 of x^2sin(3x) ?
The integral from 0 to pi/3 of x^2sin(3x) is (pi^2-4)/(27)