How do you evaluate the definite integral #int sin3x# from #[0,pi]#?

Answer 1

#2/3#

#int_0^(pi)sin3xdx#
now#d/(dx)(cos3x)=-3sin3x#
#:.int_0^(pi)sin3xdx=[-1/3cos3x]_0^pi#
#=-1/3{[cos3x]^pi-[cos3x]_0}#
#=-1/3(cos3pi-cos0)#
#=-1/3(-1-1)#
#-1/3xx-2#
#=2/3#
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Answer 2

# int_0^pi sin3x = 2/3#

Using:

# d/dx{cosax}=-asinax#

Then:

# int_0^pi sin3x = [-1/3cos3x]_0^pi # # \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -1/3(cos3pi - cos 0)# # \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = -1/3(-1- 1)# # \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2/3#
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Answer 3

To evaluate the definite integral ∫sin(3x) from 0 to π, you can use the integration by substitution method. Let u = 3x, then du = 3dx. The integral becomes ∫(1/3)sin(u) du from 0 to 3π.

Applying the antiderivative of sin(u), which is -cos(u), the integral becomes -(1/3)cos(u) evaluated from 0 to 3π.

Plugging in the upper and lower limits, you get -(1/3)cos(3π) - (-(1/3)cos(0)), which simplifies to (1/3) - (1/3) = 0.

Therefore, the value of the definite integral of sin(3x) from 0 to π is 0.

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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.

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.

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.

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.

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