Determine \[ \int_3^{10}2x^2-x+15\,dx. \]

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Lesson Parent: 
Chapter 27 - The Fundamental Theorem of Integral Calculus
Problem: 

Determine
\[
\int_3^{10}2x^2-x+15\,dx.
\]

Answer: 

$\int_3^{10}2x^2-x+15=4249/6$

Solution: 

The Fundamental Theorem of Calculus states that
\[
\int_3^{10}2x^2-x+15\,dx=F(10)-F(3)
\]
where $F(x)$ is the antiderivative of $f(x)=2x^2-x+15$. We determined $F(x)$ here and found that
\[
F(x)=\frac{2}{3}x^3-\frac{1}{2}x^2+15x+C.
\]

So now
\[
F(10)=\frac{2300}{3}+C
\]
and
\[
F(3)=\frac{117}{2}+C
\]
so that
\[
F(10)-F(3)=\frac{4249}{6}.
\]