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

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

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

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

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

Decimal

0.28049 \dots

Solution steps

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

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}{d x} ((\frac{d}{d x} (x^{3}))^{2})

\frac{d}{d x} (3 x^{2} - 4 x + 2)

\frac{d y}{d x} + y = e^{6 x}

\frac{d}{d x} (cos^{2} (sin (2 x)))

\frac{d}{d x} ((x^{2} + 1)^{l n (x)})

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