What is the integral from 0 to 1 of 6x^2cos(x^3) ?

The integral from 0 to 1 of 6x^2cos(x^3) is 2sin(1)

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integral from 0 to 1 of 6x^2cos(x^3)

\int_{0}^{1} 6 x^{2} cos (x^{3}) d x

2 sin (1)

Decimal

1.68294 \dots

Solution steps

\int_{0}^{1} 6 x^{2} cos (x^{3}) d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} x^{2} cos (x^{3}) d x

Apply u-substitution

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

= 6 \cdot \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

= 6 \cdot \frac{1}{3} \cdot \int_{0}^{1} cos (u) d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} [sin (u)]_{0}^{1}

Simplify

6 \cdot \frac{1}{3} [sin (u)]_{0}^{1} : 2 [sin (u)]_{0}^{1}

= 2 [sin (u)]_{0}^{1}

Compute the boundaries:

sin (1)

= 2 sin (1)

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What is the integral from 0 to 1 of 6x^2cos(x^3) ?
The integral from 0 to 1 of 6x^2cos(x^3) is 2sin(1)