What is the integral from-1 to 0 of-xcos(npix) ?

The integral from-1 to 0 of-xcos(npix) is -(1-(-1)^n)/(pi^2n^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from-1 to 0 of-xcos(npix)

\int_{- 1}^{0} - x cos (nπ x) d x

- \frac{1 - ( - 1 ) ^{n}}{π ^{2} n ^{2}}

Solution steps

\int_{- 1}^{0} - x cos (nπ x) d x

Take the constant out:

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

= - \int_{- 1}^{0} x cos (nπ x) d x

Apply u-substitution

: \int_{- πn}^{0} \frac{u cos ( u )}{π ^{2} n ^{2}} d u

= - \int_{- πn}^{0} \frac{u cos ( u )}{π ^{2} n ^{2}} d u

Take the constant out:

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

= - \frac{1}{π ^{2} n ^{2}} \cdot \int_{- πn}^{0} u cos (u) d u

Apply Integration By Parts

: [u sin (u) - \int sin (u) d u]_{- πn}^{0}

= - \frac{1}{π ^{2} n ^{2}} [u sin (u) - \int sin (u) d u]_{- πn}^{0}

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

= - \frac{1}{π ^{2} n ^{2}} [u sin (u) - (- cos (u))]_{- πn}^{0}

Simplify

= - \frac{1}{π ^{2} n ^{2}} [u sin (u) + cos (u)]_{- πn}^{0}

Compute the boundaries:

1 - (- 1)^{n}

= - \frac{1}{π ^{2} n ^{2}} (1 - (- 1)^{n})

Simplify

= - \frac{1 - ( - 1 ) ^{n}}{π ^{2} n ^{2}}

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

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

\int_{0}^{\frac{π}{4}} sin (θ) d θ

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

\int_{0}^{6} 2 x^{2} - x + 100 d x

What is the integral from-1 to 0 of-xcos(npix) ?
The integral from-1 to 0 of-xcos(npix) is -(1-(-1)^n)/(pi^2n^2)