What is the integral from 0 to 1 of x^4sin(pix) ?

The integral from 0 to 1 of x^4sin(pix) is (pi^4-12pi^2+48)/(pi^5)

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integral from 0 to 1 of x^4sin(pix)

\int_{0}^{1} x^{4} sin (π x) d x

\frac{π ^{4} - 12 π ^{2} + 48}{π ^{5}}

Decimal Notation

0.08814 \dots

Solution steps

\int_{0}^{1} x^{4} sin (π x) d x

Apply u-substitution

: \int_{0}^{π} \frac{u ^{4} sin ( u )}{π ^{5}} d u

= \int_{0}^{π} \frac{u ^{4} sin ( u )}{π ^{5}} d u

Take the constant out:

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

= \frac{1}{π ^{5}} \cdot \int_{0}^{π} u^{4} sin (u) d u

Apply Integration By Parts

: [- u^{4} cos (u) - \int - 4 u^{3} cos (u) d u]_{0}^{π}

= \frac{1}{π ^{5}} [- u^{4} cos (u) - \int - 4 u^{3} cos (u) d u]_{0}^{π}

\int - 4 u^{3} cos (u) d u = - 4 (u^{3} sin (u) - 3 (- u^{2} cos (u) + 2 (u sin (u) + cos (u))))

= \frac{1}{π ^{5}} [- u^{4} cos (u) - (- 4 (u^{3} sin (u) - 3 (- u^{2} cos (u) + 2 (u sin (u) + cos (u)))))]_{0}^{π}

Simplify

= \frac{1}{π ^{5}} [- u^{4} cos (u) + 4 (u^{3} sin (u) - 3 (- u^{2} cos (u) + 2 (u sin (u) + cos (u))))]_{0}^{π}

Compute the boundaries:

π^{4} - 12 π^{2} + 48

= \frac{1}{π ^{5}} (π^{4} - 12 π^{2} + 48)

Simplify

= \frac{π ^{4} - 12 π ^{2} + 48}{π ^{5}}

\int_{0}^{3} 2 π y (y^{2}) d y

\int_{0}^{1} 8 x e^{5 x} d x

\int_{0}^{4} x^{2} - 8 x + 16 d x

\int_{0}^{π} a sin (x) d x

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

What is the integral from 0 to 1 of x^4sin(pix) ?
The integral from 0 to 1 of x^4sin(pix) is (pi^4-12pi^2+48)/(pi^5)