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How do you evaluate the definite integral #int sinx dx# from #[0, pi/2]#?

Answer 1

I tried this:

First you solve the indefinite integral trying to find the Primitive (or anti-derivative) of your integrand; this is a function that DERIVED will give you #sin(x)#:

#intsin(x)dx=-cos(x)#

this is because if you derive #-cos(x)# you'll get exactly #sin(x)#!!!

After that we use the Fundamental Theorem of Calculus where, basically, we take our Primitive, we stick in first our upper extreme #pi/2# then the lower one #0# and subtract the values we obtain:

#-cos(pi/2)-(-cos(0))=1#

thus we obtain:

#int_0^(pi/2)sin(x)dx=1#

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Answer 2

Isabella Ackerman

To evaluate the definite integral ∫ sin(x) dx from 0 to π/2:

∫ sin(x) dx = -cos(x) + C

Evaluate this expression from 0 to π/2:

[-cos(π/2)] - [-cos(0)]

= -[cos(π/2)] - (-cos(0))

= -[0] - (-1)

= -0 - (-1)

= 1

So, the value of the definite integral ∫ sin(x) dx from 0 to π/2 is 1.

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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.