How do you evaluate the definite integral $int 4-x$ from $[0,2]$ ?

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Answer 1 · 2016-10-28T15:24:32

The integral $int_0^2(4-x)dx=6$

You have to use $intx^ndx=x^(n+1)/(n+1)+C$ $(n!=-1)$
$int(4-x)dx=int4dx-intxdx=4x-x^2/2$

then you have to put the limits

$int_0^2(4-x)dx=(4x-x^2/2)_0^2=(4*2-2*2/2)=6$

How do you evaluate the definite integral int 4-xint 4-x from [0,2][0,2]?