What is the integral from-1 to 2 of 12xcos(3x^2) ?

The integral from-1 to 2 of 12xcos(3x^2) is 2(sin(12)-sin(3))

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integral from-1 to 2 of 12xcos(3x^2)

\int_{- 1}^{2} 12 x cos (3 x^{2}) d x

2 (sin (12) - sin (3))

Decimal

- 1.35538 \dots

Solution steps

\int_{- 1}^{2} 12 x cos (3 x^{2}) d x

Take the constant out:

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

= 12 \cdot \int_{- 1}^{2} x cos (3 x^{2}) d x

Apply u-substitution

= 12 \cdot \int_{3}^{12} \frac{cos ( u )}{6} d u

Take the constant out:

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

= 12 \cdot \frac{1}{6} \cdot \int_{3}^{12} cos (u) d u

Use the common integral:

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

= 12 \cdot \frac{1}{6} [sin (u)]_{3}^{12}

Simplify

12 \cdot \frac{1}{6} [sin (u)]_{3}^{12} : 2 [sin (u)]_{3}^{12}

= 2 [sin (u)]_{3}^{12}

Compute the boundaries:

sin (12) - sin (3)

= 2 (sin (12) - sin (3))

What is the integral from-1 to 2 of 12xcos(3x^2) ?
The integral from-1 to 2 of 12xcos(3x^2) is 2(sin(12)-sin(3))