What is the net area between #f(x) = e^(3-x)-2x+1# and the x-axis over #x in [0, 3 ]#?

Question

1370 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-24T12:50:31

The area is #=13.09u^2#

The area is

#A=int_0^3(e^(3-x)-2x+1)dx#

#=[-e^(3-x)-x^2+x]_0^3#

#=(-1-9+3)-(-e^3-0+0)#

#=e^3-7#

#=13.09u^2#

graph{e^(3-x)-2x-1 [-28.86, 28.9, -14.44, 14.41]}

Answer 2 · 2017-07-24T12:58:49

#int_0^3(e^(3-x)-2x+1)dx=e^3-7#

Let us write #u=3-x# hence #du=-dx#,

hence #inte^(3-x)dx=-inte^udu=-e^u=-e^(3-x)#

The net area between #f(x) = e^(3-x)-2x+1# and the x-axis over #x in [0, 3 ]# is derived by

#int_0^3(e^(3-x)-2x+1)dx#

= #int_0^3e^(3-x)dx-int_0^3 2xdx+int_0^3dx#

= #[-e^(3-x)-x^2+x]_0^3#

= #(-e^0-3^2+3)-(-e^3-0^2+0)#

= #-1-9+3+e^3#

= #e^3-7#