What is the integral from 3 to 4 of 3xcos(x^2) ?

The integral from 3 to 4 of 3xcos(x^2) is 3/2 (sin(16)-sin(9))

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integral from 3 to 4 of 3xcos(x^2)

\int_{3}^{4} 3 x cos (x^{2}) d x

\frac{3}{2} (sin (16) - sin (9))

Decimal Notation

- 1.05003 \dots

Solution steps

\int_{3}^{4} 3 x cos (x^{2}) d x

Take the constant out:

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

= 3 \cdot \int_{3}^{4} x cos (x^{2}) d x

Apply u-substitution

: \int_{9}^{16} \frac{cos ( u )}{2} d u

= 3 \cdot \int_{9}^{16} \frac{cos ( u )}{2} d u

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int_{9}^{16} cos (u) d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} [sin (u)]_{9}^{16}

Simplify

3 \cdot \frac{1}{2} [sin (u)]_{9}^{16} : \frac{3}{2} [sin (u)]_{9}^{16}

= \frac{3}{2} [sin (u)]_{9}^{16}

Compute the boundaries:

sin (16) - sin (9)

= \frac{3}{2} (sin (16) - sin (9))

\int (3 x + 1)^{3} d x

- 3 x^{2}

\frac{d}{d x} (2 x^{2} - 4 x y + 3 y^{2} + 6 z^{2})

\int \frac{x - 3}{x + 3} d x

\frac{d}{d x} (\frac{1}{2} x^{4} - 4 x^{2} + 4)

What is the integral from 3 to 4 of 3xcos(x^2) ?
The integral from 3 to 4 of 3xcos(x^2) is 3/2 (sin(16)-sin(9))