# The Definite Integral

Consider an expression $F\left( x \right)$ such that
$\frac{d}{{dx}}\left[ {f\left( x \right)} \right] = 4{x^3}$

Integrating both sides with respect to $x$, we have
$\begin{gathered} \int {d\left[ {f\left( x \right)} \right]} = \int {4{x^3}dx} \\ \Rightarrow F\left( x \right) = \frac{{4{x^4}}}{4} + c \\ \Rightarrow F\left( x \right) = {x^4} + c\,\,\,\,\,{\text{ – – – }}\left( {\text{i}} \right) \\ \end{gathered}$

Putting the value $x = a$ in equation (i), we have
$F\left( a \right) = {a^4} + c$

Similarly, putting the value $x = b$ in equation (i), we have
$F\left( b \right) = {b^4} + c$

Subtracting $F\left( b \right)$ from $F\left( a \right)$, we have
$F\left( b \right) – F\left( a \right) = \left( {{b^4} + c} \right) – \left( {{a^4} + c} \right) = {b^4} – {a^4}$

We can write the above expression as
$F\left( b \right) – F\left( a \right) = \left| {{x^4}} \right|_a^b\,\,\,\,\,{\text{ – – – }}\left( {{\text{ii}}} \right)$

Since ${x^4}$ is the anti-derivative $4{x^3}$, so $\int {4{x^3}dx = {x^4}}$, where we have ignored $c$ as it is cancelled out in (ii). Making this substitution in (ii), we have
$F\left( b \right) – F\left( a \right) = \left| {\int {4{x^3}dx} } \right|_a^b$
$F\left( b \right) – F\left( a \right) = \int\limits_a^b {4{x^3}dx}$

We conclude that if $F\left( x \right)$ is the anti-derivative of $f\left( x \right)$, i.e.
$\frac{d}{{dx}}\left[ {F\left( x \right)} \right] = f\left( x \right)$

Then it can be written as
$\int\limits_a^b {f\left( x \right)dx = F\left( b \right) – F\left( a \right)}$

Thus, we have
$\frac{d}{{dx}}\left[ {f\left( x \right)} \right] = f\left( x \right) \Rightarrow \int\limits_a^b {f\left( x \right)dx = F\left( b \right) – F\left( a \right)}$

The integral $\int\limits_a^b {f\left( x \right)dx}$ is known as the definite integral.