What is the integral from 0 to pi/2 of (sin(2x)) ?

The integral from 0 to pi/2 of (sin(2x)) is 1

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integral from 0 to pi/2 of (sin(2x))

\int_{0}^{\frac{π}{2}} (sin (2 x)) d x

1

Solution steps

\int_{0}^{\frac{π}{2}} sin (2 x) d x

Apply u-substitution

: \int_{0}^{π} sin (u) \frac{1}{2} d u

= \int_{0}^{π} sin (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{π} sin (u) d u

Use the common integral:

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

= \frac{1}{2} [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= \frac{1}{2} \cdot 2

Simplify

= 1

x \to 1 - lim (2 x^{3} + 5 x)

\int_{- 2}^{5} (- 2) d x

f (x) = x^{2} + 5 x - 3

x \to 7 lim (f (x))

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

What is the integral from 0 to pi/2 of (sin(2x)) ?
The integral from 0 to pi/2 of (sin(2x)) is 1