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How do you evaluate the definite integral #int 2^x dx# from #[-1,1]#?

Answer 1

Emma Aldridge

#3/(2ln2)#

let's start by differentiating #" "2^x#.

#y=2^x=>lny=xln2#

#1/y(dy)/(dx)=ln2=>(dy)/(dx)=yln2=2^xln2#

so remembering integration is the reverse of differentiating.

#y=2^x=>(dy)/(dx)=2^xln2#

#int2^xdx=2^x/ln2+C#

#2^x>0, AA x inRR#, there is no negative area so we can

insert the limits and evaluate immediately.

#=1/ln2[2^x]_(-1)^(1)#

#=1/ln2(2^1-2^(-1))#

#=3/(2ln2)#

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Answer 2

Axel Alvarez

To evaluate the definite integral ∫2^x dx from -1 to 1, you first need to find the antiderivative of 2^x with respect to x. The antiderivative of 2^x is (1/ln(2)) * 2^x. Then, you evaluate this antiderivative at the upper and lower bounds of integration (-1 and 1) and subtract the value at the lower bound from the value at the upper bound.

So, ∫2^x dx from -1 to 1 = [(1/ln(2)) * 2^x] evaluated from -1 to 1 = [(1/ln(2)) * 2^1] - [(1/ln(2)) * 2^(-1)] = (2/ln(2)) - (1/2ln(2))

Therefore, the value of the definite integral is (2/ln(2)) - (1/2ln(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.