How do you evaluate the definite integral #int(e^x)dx# from #[-1,0]#?

Answer 1

#1-1/e#

#e^x" "# remains unchanged on either differentiating or integrating.
#int_-1^0e^xdx#
#=[e^x]_-1^0#
#e^0-e^-1#
#1-1/e#
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Answer 2

To evaluate the definite integral of ( \int_{-1}^{0} e^x , dx ), you can use the fundamental theorem of calculus. First, find the antiderivative of ( e^x ), which is ( e^x ). Then, evaluate the antiderivative at the upper and lower limits of integration and subtract the lower value from the upper value.

( \int_{-1}^{0} e^x , dx = e^x \Big|_{-1}^{0} )

( = e^0 - e^{-1} )

( = 1 - \frac{1}{e} )

So, ( \int_{-1}^{0} e^x , dx = 1 - \frac{1}{e} ).

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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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