What is the integral from 0 to 4 of sin(sqrt(x)) ?

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

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integral from 0 to 4 of sin(sqrt(x))

\int_{0}^{4} sin (x) d x

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

Decimal Notation

3.48318 \dots

Solution steps

\int_{0}^{4} sin (x) d x

Apply u-substitution

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

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

sin (2) - 2 cos (2)

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

\frac{\partial}{\partial y} (x y)

\int 3 (x^{3} - x) d x

\int \frac{2}{3} (x + 1)^{\frac{3}{2}} d x

\int_{1}^{10} 4 d x

y = x - 3

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