What is the integral from 0 to 1 of xsin((npix)/2) ?

The integral from 0 to 1 of xsin((npix)/2) is (2(2sin((pin)/2)-pincos((pin)/2)))/(pi^2n^2)

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integral from 0 to 1 of xsin((npix)/2)

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

\frac{2 ( 2 sin ( \frac{πn}{2} ) - πn cos ( \frac{πn}{2} ) )}{π ^{2} n ^{2}}

Solution steps

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

Apply u-substitution

: \int_{0}^{\frac{πn}{2}} \frac{4 u sin ( u )}{π ^{2} n ^{2}} d u

= \int_{0}^{\frac{πn}{2}} \frac{4 u sin ( 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{4}{π ^{2} n ^{2}} \cdot \int_{0}^{\frac{πn}{2}} u sin (u) d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{0}^{\frac{πn}{2}}

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - \int - cos (u) d u]_{0}^{\frac{πn}{2}}

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

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) - (- sin (u))]_{0}^{\frac{πn}{2}}

Simplify

= \frac{4}{π ^{2} n ^{2}} [- u cos (u) + sin (u)]_{0}^{\frac{πn}{2}}

Compute the boundaries:

sin (\frac{πn}{2}) - \frac{πn}{2} cos (\frac{πn}{2})

= \frac{4}{π ^{2} n ^{2}} (sin (\frac{πn}{2}) - \frac{πn}{2} cos (\frac{πn}{2}))

Simplify

= \frac{2 ( 2 sin ( \frac{πn}{2} ) - πn cos ( \frac{πn}{2} ) )}{π ^{2} n ^{2}}

x \to y lim (x - y)

x \to - 1 lim (2 - x)

f (x) = (x - 2) (3 - 2 x) (6 - x)

\int x^{2} - 4 d x

\int_{3}^{5} 3 t^{\frac{1}{2}} + 4 d t

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