What is the integral from 1 to 2 of r^7sin(2r^4) ?

The integral from 1 to 2 of r^7sin(2r^4) is 1/16 (-32cos(32)+sin(32)+2cos(2)-sin(2))

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integral from 1 to 2 of r^7sin(2r^4)

\int_{1}^{2} r^{7} sin (2 r^{4}) d r

\frac{1}{16} (- 32 cos (32) + sin (32) + 2 cos (2) - sin (2))

Decimal

- 1.74283 \dots

Solution steps

\int_{1}^{2} r^{7} sin (2 r^{4}) d r

Apply u-substitution

: \int_{2}^{32} \frac{u sin ( u )}{16} d u

= \int_{2}^{32} \frac{u sin ( u )}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int_{2}^{32} u sin (u) d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{2}^{32}

= \frac{1}{16} [- u cos (u) - \int - cos (u) d u]_{2}^{32}

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

= \frac{1}{16} [- u cos (u) - (- sin (u))]_{2}^{32}

Simplify

= \frac{1}{16} [- u cos (u) + sin (u)]_{2}^{32}

Compute the boundaries:

- 32 cos (32) + sin (32) + 2 cos (2) - sin (2)

= \frac{1}{16} (- 32 cos (32) + sin (32) + 2 cos (2) - sin (2))

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What is the integral from 1 to 2 of r^7sin(2r^4) ?
The integral from 1 to 2 of r^7sin(2r^4) is 1/16 (-32cos(32)+sin(32)+2cos(2)-sin(2))