If $\displaystyle f:[a,b]\to\mathbb R$ is an integrable function in the interval $\displaystyle [a,b]$ and if $\displaystyle \varphi:[a,b]\to\mathbb R$ is a primitive function of $\displaystyle f$ in $\displaystyle [a,b],$ then it verifies that $\displaystyle \int_a^b f(x)\,dx=\varphi(b)-\varphi(a).$

*Proof*
For any $\displaystyle P=\{x_0,x_1,\ldots,x_n\}$ partition of $\displaystyle [a,b],$ by applyin' the MVT to $\displaystyle \varphi$ in each one of intervals $\displaystyle [x_{i-1},x_i],$ we can ensure that it does exist a point $\displaystyle \mu_i\in[x_{i-1},x_i]$ (for each $\displaystyle i=1,2,\ldots,n$) such that $\displaystyle \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right)=\varphi '\left( \mu _{i} \right)\left( x_{i}-x_{i-1} \right)=f\left( \mu _{i} \right)\cdot \Delta x_{i}.$ Summing member to member the previous equalities, we get $\displaystyle \varphi (b)-\varphi (a)=\sum\limits_{i=1}^{n}{\big( \varphi \left( x_{i} \right)-\varphi \left( x_{i-1} \right) \big)}=\sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}.$

Let $\displaystyle m_i$ and $\displaystyle M_i$ be the infimum and supremum of $\displaystyle f$ in $\displaystyle [x_{i-1},x_i]$ and according to lower and upper sums $\displaystyle s(P)$ and $\displaystyle S(P)$ of $\displaystyle f$ to the $\displaystyle P$ partition, respectively, as one wants it $\displaystyle m_i=f\left( \mu _{i} \right)\le M_i,$ it yields

$\displaystyle s(P)\le \sum\limits_{i=1}^{n}{f\left( \mu _{i} \right)\cdot \Delta x_{i}}\le S(P)\implies s(P)\le \varphi (b)-\varphi (a)\le S(P).$

Hence, $\displaystyle \varphi (b)-\varphi (a)$ is included between the lower and upper sums of $\displaystyle f$ for any $\displaystyle P$ partition, and this property has the integral $\displaystyle \int_a^b f,$ from here $\displaystyle \int_a^b f(x)\,dx=\varphi(b)-\varphi(a).\quad\blacksquare$