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How do you evaluate the definite integral #int sintheta # from [0, 3pi/4]?

Answer 1

Luke Alvarez

#1/sqrt2(sqrt2-1)#

Using the standard integral.

#color(orange)"Reminder"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(intsinthetad theta=-costheta)color(white)(2/2)|)))#

#rArrint_0^((3pi)/4)sinthetad theta#

#=[-costheta]_0^((3pi)/4)#

#=(-cos((3pi)/4)-(-cos0))#

#=-1/sqrt2+1=1/sqrt2(sqrt2-1)#

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Answer 2

Aurora Alvarado

To evaluate the definite integral ∫sin(θ) from 0 to 3π/4, you use the fundamental theorem of calculus.

∫sin(θ) dθ = -cos(θ) evaluated from 0 to 3π/4.

Substituting the upper and lower limits:

= -cos(3π/4) - (-cos(0))

= -(-sqrt(2)/2) - (-1)

= sqrt(2)/2 + 1.

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Answer 3

Elijah Abram

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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.