What is the integral from-2 to 1 of 8xcos(2x^2) ?

The integral from-2 to 1 of 8xcos(2x^2) is -2(sin(8)-sin(2))

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integral from-2 to 1 of 8xcos(2x^2)

\int_{- 2}^{1} 8 x cos (2 x^{2}) d x

- 2 (sin (8) - sin (2))

Decimal

- 0.16012 \dots

Solution steps

\int_{- 2}^{1} 8 x cos (2 x^{2}) d x

Take the constant out:

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

= 8 \cdot \int_{- 2}^{1} x cos (2 x^{2}) d x

Apply u-substitution

: \int_{8}^{2} \frac{cos ( u )}{4} d u

= 8 \cdot \int_{8}^{2} \frac{cos ( u )}{4} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= 8 (- \int_{2}^{8} \frac{cos ( u )}{4} d u)

Take the constant out:

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

= 8 (- \frac{1}{4} \cdot \int_{2}^{8} cos (u) d u)

Use the common integral:

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

= 8 (- \frac{1}{4} [sin (u)]_{2}^{8})

Simplify

8 (- \frac{1}{4} [sin (u)]_{2}^{8}) : - 2 [sin (u)]_{2}^{8}

= - 2 [sin (u)]_{2}^{8}

Compute the boundaries:

sin (8) - sin (2)

= - 2 (sin (8) - sin (2))

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What is the integral from-2 to 1 of 8xcos(2x^2) ?
The integral from-2 to 1 of 8xcos(2x^2) is -2(sin(8)-sin(2))