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?

Question

Scarlett Allison · Accepted Answer

Evaluate

#int_0^π|-4sin(x)-sin(2x)|dx#

Area is: #8# The area between two continuous functions #f(x)# and #g(x)# over #x in[a,b]# is: #int_a^b|f(x)-g(x)|dx# Therefore, we must find when #f(x)>g(x)# Let the curves be the functions: #f(x)=-4sin(x)# #g(x)=sin(2x)# #f(x)>g(x)# #-4sin(x)>sin(2x)# Knowing that #sin(2x)=2sin(x)cos(x)# #-4sin(x)>2sin(x)cos(x)# Divide by #2# which is positive: #-2sin(x)>sin(x)cos(x)# Divide by #sinx# without reversing the sign, since #sinx>0# for every #x in(0,π)# #-2>cos(x)# Which is impossible, since: #-1<=cos(x)<=1# So the initial statement cannot be true. Therefore, #f(x)<=g(x)# for every #x in[0,π]# The integral is calculated: #int_a^b|f(x)-g(x)|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)# #1/2*(1-1)-4*(-1-1)# #8#

Christopher Agresta · Answer

undefined To find the area bounded by the curves $y = -4\sin(x)$ and $y = \sin(2x)$ over the closed interval from 0 to $\pi$, you need to find the points of intersection between the two curves within this interval, and then integrate the absolute difference between them with respect to $x$.

First, find the points of intersection by setting the equations equal to each other and solving for $x$.

$$
-4\sin(x) = \sin(2x)
$$

Then solve for $x$ within the interval $0 \leq x \leq \pi$.

After finding the points of intersection, integrate the absolute difference between the curves over the interval $0 \leq x \leq \pi$. The formula for the area is:

$$
\text{Area} = \int_{0}^{\pi} |\sin(2x) - (-4\sin(x))| \,dx
$$