What is the integral from 0 to 2 of x^2sin(x^3) ?

The integral from 0 to 2 of x^2sin(x^3) is 1/3 (-cos(8)+1)

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integral from 0 to 2 of x^2sin(x^3)

\int_{0}^{2} x^{2} sin (x^{3}) d x

\frac{1}{3} (- cos (8) + 1)

Decimal

0.38183 \dots

Solution steps

\int_{0}^{2} x^{2} sin (x^{3}) d x

Apply u-substitution

= \int_{0}^{8} \frac{sin ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int_{0}^{8} sin (u) d u

Use the common integral:

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

= \frac{1}{3} [- cos (u)]_{0}^{8}

Compute the boundaries:

- cos (8) + 1

= \frac{1}{3} (- cos (8) + 1)

What is the integral from 0 to 2 of x^2sin(x^3) ?
The integral from 0 to 2 of x^2sin(x^3) is 1/3 (-cos(8)+1)