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How do you compute the value of #int sint dt# of #[0, pi]#?

Answer 1

Samantha Adkison

This is a basic integral, where #intsin(t)dt=-cos(t)+C#.

Evaluating the integral on the interval #[0,pi]#:

#-cos(pi)-(-cos(0))#

#=1-(-1)=2#

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Answer 2

Paisley Johnson

#2#

#int_0^(pi)sintdt#

#=[-cost]_0^(pi)#

#=(-cospi)-(-cos0)#

#=1+1#

#=2#

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Answer 3

Scarlett Allison

To compute the integral of sin(t) with respect to t over the interval [0, π], you can use integration techniques. The integral of sin(t) is equal to -cos(t), so integrating sin(t) with respect to t gives -cos(t) + C, where C is the constant of integration. To evaluate the definite integral over the interval [0, π], you substitute the upper and lower limits of integration into the antiderivative and subtract the result. So, for the integral of sin(t) from 0 to π, you have:

cos(π) - (-cos(0))

Since cos(π) = -1 and cos(0) = 1, the integral evaluates to:

(-1) - (-1) = 1 - (-1) = 1 + 1 = 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.