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

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

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

\int_{- 1}^{0} - x sin (nπ x) d x

\frac{( - 1 ) ^{n}}{πn}

Solution steps

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

Apply Integration By Parts

: [- \frac{1}{πn} x cos (πn x) - \int - \frac{1}{πn} cos (πn x) d x]_{- 1}^{0}

= - [- \frac{1}{πn} x cos (πn x) - \int - \frac{1}{πn} cos (πn x) d x]_{- 1}^{0}

\int - \frac{1}{πn} cos (πn x) d x = - \frac{1}{π ^{2} n ^{2}} sin (πn x)

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

Simplify

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

Compute the boundaries:

- (- 1)^{n} \frac{1}{πn}

= - (- (- 1)^{n} \frac{1}{πn})

Simplify

= \frac{( - 1 ) ^{n}}{πn}

\int_{1}^{3} x^{2} - 2 x - 3 d x

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

\int_{1}^{2} 3 x^{2} + 4 x + 1 d x

\int_{0}^{2 π} (4 - cos (3 x))^{2} d x

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