How do you evaluate the definite integral #int(-2x+2)dx# from #[-2,3]#?

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Answer 1 · 2016-11-03T19:20:16

I found: #int_-2^3(-2x+2)dx=5#

We can write the integral as:

#int-2xdx+int2dx=-2intxdx+2intdx=#

we can now integrate to get:

#=-2x^2/2+2x=-x^2+2x#

we now substitute the exteremes of integration into our result and subtract:

for #x=3# we get::#-9+6=-3#

for #x=-2# we get: #-4-4=-8#

Let us subtract them:
#-3-(-8)=5#
so finally:

#int_-2^3(-2x+2)dx=5#