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

Feb 10, 2015

You can integrate separately each function and then, once you found the anti-derivative $F \left(x\right)$, substitute $0$ and $7$:
${\int}_{0}^{7} f \left(x\right) \mathrm{dx} = F \left(7\right) - F \left(0\right)$
${\int}_{0}^{7} \left(3 {x}^{2} - 5 x + 9\right) \mathrm{dx} =$
$= {\int}_{0}^{7} \left(3 {x}^{2}\right) \mathrm{dx} - {\int}_{0}^{7} \left(5 x\right) \mathrm{dx} + {\int}_{0}^{7} \left(9\right) \mathrm{dx} =$
Remember that $\int k {x}^{n} \mathrm{dx} = k {x}^{n + 1} / \left(n + 1\right) + c$ where $k$ is a constant and $\int k \mathrm{dx} = k x + c$;
=3x^3/3-5x^2/2+9x]_0^7= you can now substitute the extremes of integration:
$= \left({7}^{3} - 5 \cdot {7}^{2} / 2 + 9 \cdot 7\right) - \left(0 - 0 + 0\right) =$
$= 343 - 122.5 + 63 = 283.5$