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

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Answer 1 · 2017-01-19T02:33:29

#int_2^4xdx=6#

#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#