You can integrate separately each function and then, once you found the anti-derivative F(x)F(x), substitute 00 and 77:
int_0^7f(x)dx=F(7)-F(0)∫70f(x)dx=F(7)−F(0)
In your case:
int_0^7(3x^2-5x+9)dx=∫70(3x2−5x+9)dx=
=int_0^7(3x^2)dx-int_0^7(5x)dx+int_0^7(9)dx==∫70(3x2)dx−∫70(5x)dx+∫70(9)dx=
Remember that intkx^ndx=kx^(n+1)/(n+1)+c∫kxndx=kxn+1n+1+c where kk is a constant and intkdx=kx+c∫kdx=kx+c;
you have:
=3x^3/3-5x^2/2+9x]_0^7==3x33−5x22+9x]70= you can now substitute the extremes of integration:
=(7^3-5*7^2/2+9*7)-(0-0+0)==(73−5⋅722+9⋅7)−(0−0+0)=
=343-122.5+63=283.5