What is the integral from 0 to 1 of 3sin(npix) ?

The integral from 0 to 1 of 3sin(npix) is (3(-(-1)^n+1))/(pin)

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integral from 0 to 1 of 3sin(npix)

\int_{0}^{1} 3 sin (nπ x) d x

\frac{3 ( - ( - 1 ) ^{n} + 1 )}{πn}

Solution steps

\int_{0}^{1} 3 sin (nπ x) d x

Take the constant out:

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

= 3 \cdot \int_{0}^{1} sin (nπ x) d x

Apply u-substitution

: \int_{0}^{πn} \frac{sin ( u )}{πn} d u

= 3 \cdot \int_{0}^{πn} \frac{sin ( u )}{πn} d u

Take the constant out:

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

= 3 \cdot \frac{1}{πn} \cdot \int_{0}^{πn} sin (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{πn} [- cos (u)]_{0}^{πn}

Simplify

3 \cdot \frac{1}{πn} [- cos (u)]_{0}^{πn} : \frac{3}{πn} [- cos (u)]_{0}^{πn}

= \frac{3}{πn} [- cos (u)]_{0}^{πn}

Compute the boundaries:

- (- 1)^{n} + 1

= \frac{3}{πn} (- (- 1)^{n} + 1)

Simplify

= \frac{3 ( - ( - 1 ) ^{n} + 1 )}{πn}

\int_{0}^{- 1} \frac{3}{1 + t ^{2}} d t

\int_{1}^{e^{x}} 3 ln (x) d x

\int_{1}^{4} e^{- x} d x

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

\int_{- 2}^{2} - x^{2} + 5 d x

What is the integral from 0 to 1 of 3sin(npix) ?
The integral from 0 to 1 of 3sin(npix) is (3(-(-1)^n+1))/(pin)