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How do you integrate #int e^(3-x)dx# from #[3,4]#?

Answer 1

Samantha Adkison

#int_3^4(e^(3-x))dx=1- 1/e#

#int_3^4(e^(3-x))dx#

Do a #u#-substitution: #u=3-x# #(du)/dx=-1# #dx=-du#

Remember to change the bounds of the integral according to the #u# values #int_(0)^(-1)(e^u)(-1)du#

#int_0^(-1)(-e^u)du#

#=[-e^u]_0^(-1)#

#=-e^(-1)+1#

#=1- 1/e#

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Answer 2

Charles Anctil

To integrate ( \int_{3}^{4} e^{3-x} , dx ), you can use the substitution method. Let ( u = 3 - x ), then ( du = -dx ). When ( x = 3 ), ( u = 0 ), and when ( x = 4 ), ( u = -1 ). Therefore, the integral becomes ( -\int_{0}^{-1} e^{u} , du ). Integrating ( e^u ) yields ( -e^u ). Evaluate ( -e^u ) from ( 0 ) to ( -1 ) to get ( -e^{-1} + e^0 ). Simplify to get the final result: ( 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.