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How do you find the definite integral for: #x dx# for the intervals #[2, 4]#?

Answer 1

Cameron Alexander

#int_2^4xdx=6#

The fundamental Theorem of Calculus says,

#int_a^bf(x)dx=F(b)-F(a)# where #F'(x)=f(x)#

In this case #f(x)=x#, #a=2#, and #b=4#

Since #((x^2)/2)'=(1/2)(2)x^(2-1)=(2/2)x^1=(1)x=x#

#F(x)=(x^2)/2#

Then

#ul(int_2^4xdx=(4^2)/2-(2^2)/2=16/2-4/2=8-2=6#

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Answer 2

Ezra Adams

To find the definite integral of ( x , dx ) over the interval ([2, 4]), you use the fundamental theorem of calculus. First, you integrate (x) with respect to (x) to get ( \frac{x^2}{2} ). Then, you evaluate this antiderivative at the upper and lower limits of integration and subtract the result of the lower limit from the result of the upper limit.

So, integrating ( x , dx ) gives ( \frac{x^2}{2} ). Evaluating this from 2 to 4:

[ \left[ \frac{x^2}{2} \right]_{2}^{4} = \frac{4^2}{2} - \frac{2^2}{2} = 8 - 2 = 6 ]

Therefore, the definite integral of ( x , dx ) over the interval ([2, 4]) is 6.

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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.