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 a definite integral.

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