Evaluate the following term #int_0^(3pi/2) 5|sinx|dx# .How would i do this using FTC2(F(b)-F(a))?

Question

Charles Arocha · Accepted Answer

We need to split up the integral.

Recall that #absu = {(u,"if",u >= 0),(-u,"if",u<0):}# so #abs(sinx) = {(sinx,"if",sinx >= 0),(-sinx,"if",sinx < 0):}#. We are integrating on #[0,(3pi)/2]# and we know that #{(sinx >=0,"if" ,0 <=x <= pi),(sinx < 0,"if",pi < x <= (3pi)/2):}# Therefore, #abs(sinx) = {(sinx,"if",0 <= x <= pi),(-sinx,"if",pi < x < (3pi)/2):}#. #int_0^((3pi)/2) 5abs(sinx) dx = 5int_0^((3pi)/2) abs(sinx) dx # # = 5[int_0^pi sinx dx + int_pi^((3pi)/2) -sinx dx]# # = 5[int_0^pi sinx dx - int_pi^((3pi)/2) sinx dx]# Now use the fact that #int sinx dx = -cosx +C# to find each of the integrals. # = 5[{:-cosx]_0^pi +{:cosx]_pi^((3pi)/2)]# # = 5[(-cospi+cos0)+(cos((3pi)/2)-cospi)]# # = 5[(-(-1)+1+0-(-1)]# # = 5[3] = 15# Bonus method Some people prefer to integrate #abs(f(x))# by simply integrating from one zero to the next without first adjusting the sign. Any integral that comes out negative, we make positive. The notation for this technique is #int_0^((3pi)/2) 5abs(sinx) dx = abs(int_0^pi 5sinx dx)+abs(int_pi^((3pi)/2) 5sinx dx)# The first of these two integrals will be positive and the second will be negative. (That's why the first method changed the sign for the second integral before integrating.)

Paisley Johnson · Answer

undefined To evaluate the integral $ \int_0^{\frac{3\pi}{2}} 5|\sin(x)| \, dx $ using the Second Fundamental Theorem of Calculus (FTC2), we first need to find the antiderivative of the integrand. Since the integrand involves the absolute value of sine, we consider two cases:

1. For $ 0 \leq x \leq \frac{\pi}{2} $, $ |\sin(x)| = \sin(x) $.
2. For $ \frac{\pi}{2} \leq x \leq \frac{3\pi}{2} $, $ |\sin(x)| = -\sin(x) $.

Let's calculate the antiderivative separately for these intervals:

1. For $ 0 \leq x \leq \frac{\pi}{2} $:
$$ \int \sin(x) \, dx = -\cos(x) + C $$

2. For $ \frac{\pi}{2} \leq x \leq \frac{3\pi}{2} $:
$$ \int -\sin(x) \, dx = \cos(x) + C $$

Now, we can use FTC2 to evaluate the integral:

$$ \int_0^{\frac{3\pi}{2}} 5|\sin(x)| \, dx = \left[ 5(-\cos(x)) \right]_0^{\frac{\pi}{2}} + \left[ 5(\cos(x)) \right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} $$

Simplifying further:
$$ = 5(-\cos(\frac{\pi}{2})) + 5(\cos(\frac{3\pi}{2})) - 5(\cos(\frac{\pi}{2})) + 5(\cos(\frac{\pi}{2})) $$

Since $ \cos(\frac{\pi}{2}) = 0 $ and $ \cos(\frac{3\pi}{2}) = 0 $, the integral simplifies to:
$$ = 5(0) + 5(0) - 5(0) + 5(0) = 0 $$

Therefore, $ \int_0^{\frac{3\pi}{2}} 5|\sin(x)| \, dx = 0 $ when evaluated using FTC2.