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What is the integral of #sqrt(cos^(2)x)# from zero to pi?

Answer 1

#int_0^ pi sqrt cos^2 xdx#= #int_0^pi cos x dx#

This definite integral means the area bounded by the curve y= cos x between x=0 to x= #pi#. To evaluate this area correctly, divide the interval in two parts 0 to #pi/2# and #pi/2 # to #pi# = #[sinx]_0^(pi/2)# +

#[sin x]_(pi/2)^(pi)#

The area above the x-axis would become +1, and the area below the x-axis would become -1. The total of the two areas would not equal 0. Rather, it would equal 1+1 = 2.

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Answer 2

Scarlett Allison

Note that #sqrt(u^2) = absu# if we cannot be sure that #u# is positive.

In the interval #[0, pi]#, we have:.

#sqrt(cos^2x ) = abs(cosx) = { (cosx, ", if " 0<= x <= pi/2), (-cosx, ", if " pi/2<= x <= pi) :}#

Thus, the integral takes on the following form:

#int_0^pi sqrt(cos^2x ) dx = int_0^pi abs(cosx) dx#

#color(white)"sssssssssssssss"# # = int_0^(pi/2) cosx dx+int_(pi/2)^pi -cosx dx#

Both of these integrals evaluate to #1#, so we get:

#int_0^pi sqrt(cos^2x ) dx = 1+1=2#

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Answer 3

Paisley Johnson

The integral of √(cos^2(x)) from 0 to π is equal to π/2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.