How do you evaluate the definite integral #int dx/x# from #[1/e,e]#?

Answer 1

2

#int_(1/e)^e(dx)/x=[lnx]_(1/e)^e#
#=lne-ln(1/e)#
#=lne-[ln1-lne]#
#=1-0+1#
#=2#
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Answer 2

The answer is #2#

The integral is #intdx/x=lnx + C#
So the integral is #intdx/x=(lnx)#
So from #(1/e,e)#, we have #intdx/x=lne-ln(1/e)=lne-ln1+lne#
As #lne=1# and #ln1=0# So we have finally #intdx/x=1-0+1=2#
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Answer 3

To evaluate the definite integral ∫ dx/x from 1/e to e, we can use the natural logarithm function. The integral of dx/x is ln|x| + C, where C is the constant of integration. Applying this formula, we get:

∫ dx/x = ln|x| + C

Now, evaluate this expression at the upper and lower limits:

∫ dx/x from 1/e to e = [ln(e) - ln(1/e)]

Using the property of logarithms, ln(e) = 1 and ln(1/e) = -1, so:

∫ dx/x from 1/e to e = (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.

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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