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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