What is the integral from 0 to 1 of s^2cos(s^3) ?

The integral from 0 to 1 of s^2cos(s^3) is 1/3 sin(1)

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integral from 0 to 1 of s^2cos(s^3)

\int_{0}^{1} s^{2} cos (s^{3}) d s

\frac{1}{3} sin (1)

Decimal Notation

0.28049 \dots

Solution steps

\int_{0}^{1} s^{2} cos (s^{3}) d s

Apply u-substitution

: \int_{0}^{1} \frac{cos ( u )}{3} d u

= \int_{0}^{1} \frac{cos ( 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}^{1} cos (u) d u

Use the common integral:

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

= \frac{1}{3} [sin (u)]_{0}^{1}

Compute the boundaries:

sin (1)

= \frac{1}{3} sin (1)

\frac{d y}{d x}, 2 x y = 9

\frac{d}{d x} (tan (x) cos (x))

\frac{d}{d x} (e^{x + 1} + 4)

\frac{\partial}{\partial x} (tan (7 + 2 x^{2} y^{4} z^{2}))

1 - e^{c o s (1 - x^{2})}

What is the integral from 0 to 1 of s^2cos(s^3) ?
The integral from 0 to 1 of s^2cos(s^3) is 1/3 sin(1)