How do you integrate #3x^2-5x+9# from 0 to 7?

1 Answer
Gió
Feb 10, 2015

You can integrate separately each function and then, once you found the anti-derivative #F(x)#, substitute #0# and #7#:
#int_0^7f(x)dx=F(7)-F(0)#
In your case:
#int_0^7(3x^2-5x+9)dx=#
#=int_0^7(3x^2)dx-int_0^7(5x)dx+int_0^7(9)dx=#

Remember that #intkx^ndx=kx^(n+1)/(n+1)+c# where #k# is a constant and #intkdx=kx+c#;

you have:

#=3x^3/3-5x^2/2+9x]_0^7=# you can now substitute the extremes of integration:

#=(7^3-5*7^2/2+9*7)-(0-0+0)=#
#=343-122.5+63=283.5#

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