What is the definite integral of $3 sin 2 x$ $3 sin2x$ from $x = 0$ $x = 0$ to $x = 2$ $x =2$ ?

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Answer 1 · 2014-09-25T04:44:29

$\int_{0}^{2} 3 sin (2 x) d x$ $int_0^2 3sin(2x)dx$

Factor out the constant

$3 \int_{0}^{2} sin (2 x) d x$ $3int_0^2 sin(2x)dx$

Make a substitution

Let $u = 2 x$ $u=2x$

$d u = 2 d x$ $du=2 dx$

$\frac{d u}{2} = \frac{2 d x}{2}$ $(du)/2=(2 dx)/2$

$\frac{d u}{2} = d x$ $(du)/2=dx$

$3 \int sin (u) \cdot \frac{d u}{2}$ $3int sin(u)*(du)/2$

$= 3 \cdot \frac{1}{2} \int sin (u) d u$ $=3*1/2int sin(u)du$

$= \frac{3}{2} \int sin (u) d u$ $=3/2int sin(u)du$

Recalculate the boundaries

Upper Bound

$u = 2 x = 2 (2) = 4$ $u=2x=2(2)=4$

Lower Bound

$u = 2 x = 2 (0) = 0$ $u=2x=2(0)=0$

$\frac{3}{2} \int_{0}^{4} sin (u) d u$ $3/2int_0^4 sin(u)du$

$= \frac{3}{2} {[- cos (u)]}_{0}^{4}$ $=3/2[-cos(u)]_0^4$

$= \frac{3}{2} [- cos (4) - (- cos (0))]$ $=3/2[-cos(4)-(-cos(0))]$

$= \frac{3}{2} [- cos (4) - (- 1)]$ $=3/2[-cos(4)-(-1)]$

$= \frac{3}{2} [- cos (4) + 1]$ $=3/2[-cos(4)+1]$

$= [\frac{- 3 cos (4) + 3}{2}]$ $=[(-3cos(4)+3)/2]$

$= 2.480465431$ $=2.480465431$ sq units

What is the definite integral of 3sin2x3 sin2x from x=0x = 0 to x=2x =2?