What is the integral from 0 to 1 of x^2cos(npix) ?

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

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integral from 0 to 1 of x^2cos(npix)

\int_{0}^{1} x^{2} cos (nπ x) d x

\frac{2 ( - 1 ) ^{n}}{π ^{2} n ^{2}}

Solution steps

\int_{0}^{1} x^{2} cos (nπ x) d x

Apply u-substitution

: \int_{0}^{πn} \frac{u ^{2} cos ( u )}{π ^{3} n ^{3}} d u

= \int_{0}^{πn} \frac{u ^{2} cos ( u )}{π ^{3} n ^{3}} d u

Take the constant out:

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

= \frac{1}{π ^{3} n ^{3}} \cdot \int_{0}^{πn} u^{2} cos (u) d u

Apply Integration By Parts

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

= \frac{1}{π ^{3} n ^{3}} [u^{2} sin (u) - \int 2 u sin (u) d u]_{0}^{πn}

\int 2 u sin (u) d u = 2 (- u cos (u) + sin (u))

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

Compute the boundaries:

2 π (- 1)^{n} n

= \frac{1}{π ^{3} n ^{3}} \cdot 2 π (- 1)^{n} n

Simplify

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

\int_{0}^{1} x^{3} 1 + x^{2} d x

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

\int_{a}^{b} cos (x) d x

\int_{y}^{1} \frac{1}{x} d x

\int_{0}^{π} (sin (x) - 2) d x

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