What is the integral from 0 to 16 of sin(sqrt(x)) ?

The integral from 0 to 16 of sin(sqrt(x)) is 2(sin(4)-4cos(4))

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integral from 0 to 16 of sin(sqrt(x))

\int_{0}^{16} sin (x) d x

2 (sin (4) - 4 cos (4))

Decimal Notation

3.71554 \dots

Solution steps

\int_{0}^{16} sin (x) d x

Apply u-substitution

: \int_{0}^{4} sin (u) \cdot 2 u d u

= \int_{0}^{4} sin (u) \cdot 2 u d u

Take the constant out:

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

= 2 \cdot \int_{0}^{4} sin (u) u d u

Apply Integration By Parts

: [- u cos (u) - \int - cos (u) d u]_{0}^{4}

= 2 [- u cos (u) - \int - cos (u) d u]_{0}^{4}

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

= 2 [- u cos (u) - (- sin (u))]_{0}^{4}

Simplify

= 2 [- u cos (u) + sin (u)]_{0}^{4}

Compute the boundaries:

sin (4) - 4 cos (4)

= 2 (sin (4) - 4 cos (4))

\int sin^{n} (x) d x

\frac{d}{d x} (x^{9 s i n (x)})

\int z 4 z^{2} + 8 d z

\frac{d y}{d x} = \frac{( x - 4 )}{y - 6}

y = 10 + 9 x - x^{2}, y = - 3 x + 46, [6, 15 + \frac{1}{3}]

What is the integral from 0 to 16 of sin(sqrt(x)) ?
The integral from 0 to 16 of sin(sqrt(x)) is 2(sin(4)-4cos(4))