# How do you find the area bounded by the curves y = -4sin(x) and y = sin(2x) over the closed interval from 0 to pi?

Mar 15, 2016

Evaluate

int_0^π|-4sin(x)-sin(2x)|dx

Area is: $8$

#### Explanation:

The area between two continuous functions $f \left(x\right)$ and $g \left(x\right)$ over $x \in \left[a , b\right]$ is:

${\int}_{a}^{b} | f \left(x\right) - g \left(x\right) | \mathrm{dx}$

Therefore, we must find when $f \left(x\right) > g \left(x\right)$

Let the curves be the functions:

$f \left(x\right) = - 4 \sin \left(x\right)$

$g \left(x\right) = \sin \left(2 x\right)$

$f \left(x\right) > g \left(x\right)$

$- 4 \sin \left(x\right) > \sin \left(2 x\right)$

Knowing that $\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

$- 4 \sin \left(x\right) > 2 \sin \left(x\right) \cos \left(x\right)$

Divide by $2$ which is positive:

$- 2 \sin \left(x\right) > \sin \left(x\right) \cos \left(x\right)$

Divide by $\sin x$ without reversing the sign, since $\sin x > 0$ for every x in(0,π)

$- 2 > \cos \left(x\right)$

Which is impossible, since:

$- 1 \le \cos \left(x\right) \le 1$

So the initial statement cannot be true. Therefore, $f \left(x\right) \le g \left(x\right)$ for every x in[0,π]

The integral is calculated:

${\int}_{a}^{b} | f \left(x\right) - g \left(x\right) | \mathrm{dx}$

int_0^π(g(x)-f(x))dx

int_0^π(sin(2x)-(-4sin(x)))dx

int_0^π(sin(2x)+4sin(x))dx

int_0^πsin(2x)dx+4int_0^πsin(x)

-1/2[cos(2x)]_0^π-4[cos(x)]_0^π

-1/2(cos2π-cos0)-4(cosπ-cos0)

$\frac{1}{2} \cdot \left(1 - 1\right) - 4 \cdot \left(- 1 - 1\right)$

$8$